Answer:
[tex] \huge{ \boxed {\bold{ \tt{160 \: N}}}}[/tex]
Explanation:
Given :Mass ( m ) = 10 kgAcceleration ( a ) = 16 m / s²To find :Force = ?We know,[tex] \boxed{ \sf{Force = Mass \: \times \: Acceleration}}[/tex]
[tex] \sf{➛ force \: = 10\: \times 16}[/tex]
[tex] \sf{➛ force = 160 \: Newton}[/tex]
The amount of force is 160 N
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A ball is thrown straight downward with a speed of 0.50 meter per second from a height of 4.0 meters. What is the speed of the ball 0.70 second after it is released? [Neglect friction.]
Answer:
The speed of the ball is 7.36 m/s
Explanation:
Constant Acceleration Motion
It happens when the velocity of an object changes by an equal amount in every equal period of time.
Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:
[tex]v_f=v_o+at[/tex]
Since the situation corresponds to a vertical motion, the acceleration is g=9.8 m/s^2.
The ball is thrown downward at vo=0.5 m/s. Its speed at t=0.7 seconds is
[tex]v_f=0.5+9.8\cdot 0.7[/tex]
[tex]v_f=7.36\ m/s[/tex]
The speed of the ball is 7.36 m/s
Note: There is no use of the height it was released from
A string under a tension of 53.9 N is used to whirl a rock in a horizontal circle of radius 2.59 m at a speed of 21.8 m/s. The string is pulled in and the speed of the rock increases. When the string is 1.23 m long and the speed of the rock is 54.8 m/s, the string breaks. What is the breaking strength (in newtons) of the string
Answer:
The breaking strength of the string is 717.19 N
Explanation:
Given;
initial tension on the string, T₁ = 53.9 N
initial radius of the string, r₁ = 2.59 m
initial speed of the rock, v₁ = 21.8 m/s
final radius of the string, r₂ = 1.23 m
final speed of the rock, v₂ = 54.8 m/s
the final tension on the string = breaking strength, T₂ = ?
Apply centripetal force equation due to inward motion;
[tex]T= \frac{mv^2}{r}\\\\Tr = mv^2\\\\m =\frac{Tr}{v^2}\\\\ \frac{T_1r_1}{v_1^2} = \frac{T_2r_2}{v_2^2}\\\\T_2 = \frac{T_1r_1v_2^2}{r_2v_1^2}\\\\ T_2 = \frac{(53.9)(2.59)(54.8)^2}{(1.23)(21.8)^2}\\\\ T_2 = 717.19 \ N[/tex]
Therefore, the breaking strength of the string is 717.19 N
A 100-g bullet is fired from a rifle having a barrel 0.600 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is 15 000 1 10 000x 2 25 000x2, where x is in meters. (a) Determine the work done by the gas on the bullet as the bullet travels the length of the barrel. (b) What If
Answer:
9000 J
Explanation:
weight of bullet = 100 g
length of gun barrel = 0.6m
A) determine the work done by the gas on the bullet as the bullet travels the length of the barrel
The work done is an integral of force over distance in this scenario
lets represent the force using some constants ( arbitrary )
Fx = a + bx + cx^2
where : a = 15000, b = 10000, c = - 25000
hence work done ( W ) = [tex]ax_{f } + \frac{bx^{2} _{f} }{2} + \frac{cx^{3} _{f} }{3}[/tex]
where ; xf = 0.6
hence work done(W)
= 15000(0.6) + [(10000*0.6^2)/2] + [(-25000*0.6^3)/3]
= 9000 J
A big dog B and a small dog S are pulling horizontally in opposite directions on a large bone. The bone’s mass is mmm, the big dog’s leftward force magnitude is F_BF
B
F, start subscript, B, end subscript and the small dog’s rightward force magnitude is F_SF
S
F, start subscript, S, end subscript. The bone accelerates to the right along the table with no friction.
The correct expression for the mass m of the bone is m=Fs-Fb/a.
What is force?The force is defined as the shear stress or pressure applied per unit area.
A big dog B and a small dog S are pulling horizontally in opposite directions on a large bone. The bone’s mass is m, the big dog’s leftward force magnitude is Fb and the small dog’s rightward force magnitude is Fs. The bone accelerates to the right along the table with no friction.
Across the bottom is green grass. In the center, there is a brown table with a large bone on top. To the left is a black outline of a large dog facing rightwards that is pulling leftwards on the bone. To the right is a black outline of a small dog facing leftwards that is pulling rightwards on the bone. The small dog stands on a grey box. Across the bottom is green grass.
Two dogs pulling on a large bone. Big dog and small dog images courtesy of pixabay. Consider rightward as the positive direction.
The net force acting on bone will be
F = Fs -Fb
From the Newton's second law of motion, Force = mass x acceleration
F = ma
m = F/a
m = Fs-Fb/a
Thus, correct expression for the mass m of the bone is m=Fs-Fb/a.
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A 3.0-kg mass is located at (0.0 m, 8.0 m), and a 1.0-kg mass is located at (12 m, 0.0 m) where (x, y) represents x and y coordinates respectively on x-y plane. You want to add a 4.0-kg mass so that the center of mass (or center of gravity) of the three-mass system will be at the origin. What should be the coordinates of the 3.0-kg mass
Answer:
The coordinates of the 4.0-kg mass must be [tex](x,y) = (-3\,m,-6\,m)[/tex].
Explanation:
We have a system formed by three particles with known masses and locations with respect to origin. The location of the center of mass is calculated by means of weighted means. That is:
[tex]\bar x = \frac{\Sigma\limits_{i=1}^{n}\,x_{i}\cdot m_{i}}{\Sigma\limits_{i=1}^{n}m_{i}}[/tex] (Eq. 1)
[tex]\bar y = \frac{\Sigma\limits_{i=1}^{n}\,y_{i}\cdot m_{i}}{\Sigma\limits_{i=1}^{n}m_{i}}[/tex] (Eq. 2)
Where:
[tex]\bar x[/tex], [tex]\bar y[/tex] - Coordinates of the center of mass of the system, measured in meters.
[tex]x_{i}[/tex], [tex]y_{i}[/tex] - Coordinates of the i-th particle, measured in meters.
[tex]m_{i}[/tex] - Mass of the i-th particle, measured in kilograms.
If we know that [tex]m_{1} = 3\,kg[/tex], [tex]m_{2} = 1\,kg[/tex], [tex]m_{3} = 4\,kg[/tex], [tex]x_{1} = 0\,m[/tex], [tex]x_{2} = 12\,m[/tex], [tex]\bar x = 0\,m[/tex], [tex]y_{1} = 8\,m[/tex], [tex]y_{2} = 0\,m[/tex] and [tex]\bar y = 0\,m[/tex], the location of the 4.0 kg mass is:
[tex]\frac{(3\,kg)\cdot (0\,m)+(1\,kg)\cdot (12\,m)+(4\,kg)\cdot x_{3}}{3\,kg+1\,kg+4\,kg} = 0\,m[/tex]
[tex]12+4\cdot x_{3} = 0[/tex]
[tex]x_{3} = -3\,m[/tex]
[tex]\frac{(3\,kg)\cdot (8\,m)+(1\,kg)\cdot (0\,m)+(4\,kg)\cdot y_{3}}{3\,kg+1\,kg+4\,kg}= 0\,m[/tex]
[tex]24+4\cdot y_{3} = 0[/tex]
[tex]y_{3} = -6\,m[/tex]
The coordinates of the 4.0-kg mass must be [tex](x,y) = (-3\,m,-6\,m)[/tex].
The coordinates of the 4kg mass so that the center of mass of the three-mass system will be at the origin is (-3,-6)
Finding the coordinates:The coordinates of the 3kg mass are (0.0 m, 8.0 m)
and the coordinates for the 1kg mass are (12 m, 0.0 m)
Let the coordinates of the 4kg mass be (x,y)
The final coordinates of the center of mass of the system are (0,0),
so the equation for x-coordinate is given by:
[tex]0=\frac{3\times0+1\times12+4\times x}{3+1+4}[/tex]
12 + 4x = 0
x = -3
the equation for y-coordinate is given by:
[tex]0=\frac{3\times8+1\times0+4\times y}{3+1+4}[/tex]
24 + 4y = 0
y = -6
So, the coordinates of the 4kg mass are (-3,-6)
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When the momentum of an object increases with respect to time, what is true of the net force acting on it?
a. It is zero, because the net force is equal to the rate of change of the momentum.
b. It is zero, because the net force is equal to the product of the momentum and the time interval.
c. It is nonzero, because the net force is equal to the rate of change of the momentum.
d. It is nonzero, because the net force is equal to the product of the momentum and the time interval.
Answer:
C. It is nonzero, because the net force is equal to the rate of change of the momentum
Explanation:
Momentum of an object is given as;
P = mv = ft
f = mv/t
or f = P/t
where;
P is the momentum of the object
m is mass of the object
v is velocity of the object
f is the applied force on the object
t is time
From the given equation above; the net force acting on the object is equal to the rate of change of the momentum, thus it is non-zero.
Therefore, the correct option is "C"
C. It is nonzero, because the net force is equal to the rate of change of the momentum.
The net force acting on the object is non-zero, because the net force is equal to the rate of change of the momentum. Hence, option (c) is correct.
The problem is based on the impulse-momentum concept. As per the impulse -momentum concept, the magnitude of impulse produced by an object is equal to the change in momentum. Then the expression is,
[tex]I = \Delta P[/tex]
here,
I is the magnitude of impulse. And its value is,
[tex]I = F \times t[/tex]
t is the impact time or time interval.
F is the magnitude of net force.
[tex]\Delta P[/tex] is the change in momentum.
Solving as,
[tex]I= \Delta P\\\\F \times t= \Delta P\\\\F = \dfrac{\Delta P}{t}[/tex]
Clearly, the net force is dependent of the change in momentum with respect to time. So, on increasing the momentum change with respect to time, the net force will also increase with non-zero value.
Thus, we can conclude that the net force acting on the object is non-zero, because the net force is equal to the rate of change of the momentum. Hence, option (c) is correct.
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An object is moving with an initial velocity of 20 m/s. If it is decelerating at 5 m/s2 how far does it travel before it stops
Answer:
The object travels a distance of 40 meters before it stops.
Explanation:
A body moves with constant acceleration motion or uniformly accelerated rectilinear motion (u.a.r.m) when the trajectory of the mobile is a straight line and its speed varies the same amount in each unit of time.
Torricelli's equation relates the speed to the space traveled by the body in u.a.r.m. To calculate the velocity of a body as a function of space, the expression is used:
[tex]v^{2} =v0^{2} +2*a*d[/tex]
Where :
v: final velocity (m/s) v0: initial velocity (m/s) a: acceleration (m/s²) d: space traveled by the body (m)In this case:
v: 0 m/s because the object stopsv0: 20 m/sa: - 5 m/s² because the object is deceleratingd: ?Replacing:
[tex](0\frac{m}{s} )^{2} =(20 \frac{m}{s} )^{2} +2*(-5\frac{m}{s^{2} }) *d[/tex]
Solving:
[tex]0\frac{m^{2} }{s^{2} } =400 \frac{m^{2} }{s^{2} } -10\frac{m}{s^{2} } *d[/tex]
[tex]0\frac{m^{2} }{s^{2} } -400 \frac{m^{2} }{s^{2} } = -10\frac{m}{s^{2} } *d[/tex]
[tex]-400 \frac{m^{2} }{s^{2} } = -10\frac{m}{s^{2} } *d[/tex]
[tex]\frac{ -400 \frac{m^{2} }{s^{2} }}{ -10\frac{m}{s^{2} }} = d[/tex]
40 m= d
The object travels a distance of 40 meters before it stops.
Calculate the linear acceleration of a car, the 0.300-m radius tires of which have an angular acceleration of 17.0 rad/s2. Assume no slippage and give your answer in m/s2.
Answer:
5.1m/s^2Explanation:
In this problem we want to solve for the linear acceleration
Given data
radius r= 0.3m
angular acceleration α=17rad/s^2
we know that the expression relating angular acceleration and linear acceleration is given as
a=rα
Substituting we have
a=0.3*17
a=5.1m/s^2
What happens if atoms lose energy during a change of state?
Answer: they become slower.
Explanation: because they get pulled together.
Answer:
The atoms are pulled together by attractive forces and become more organized.
Explanation:
Atoms lose energy as a gas changes to a solid. ... Atoms gain energy as a solid changes to a liquid. If atoms energy during a change of state, they are pulled together by attractive forces and become more organized.
Reference by Quizlet.
If we put a flammable material in a point where many light rays converge we can set it on fire. With what type of spherical mirror would it be possible to do this
Answer:
Concave mirror
Explanation:
A concave mirror is a type that has the property of converging rays of light that are parallel to the principal axis to a point of focus. Thus, for the convergence of many rays of light (i.e a beam of light) as required in the given question, a concave mirror would serve the purpose. Curved mirrors have various applications in optics.
A small, 300 gg cart is moving at 1.30 m/sm/s on a frictionless track when it collides with a larger, 4.00 kgkg cart at rest. After the collision, the small cart recoils at 0.860 m/sm/s . Part A What is the speed of the large cart after the collision
Answer:
v₂f = 0.16 m/s (to the right)
Explanation:
Assuming no external forces acting during the collision, as stated by Newton's 2nd law, total momentum must be conserved.The initial momentum is due to the small cart only, because the big one is at rest.Assuming that the small cart is going to the right and taking this direction as the positive one, we can write the following expression for the initial momentum p₀ :[tex]p_{o} = m_{1} * v_{o1} = 0.3 kg * 1.3 m/s (1)[/tex]
The final momentum can be written as follows:[tex]p_{f} = m_{1} * v_{1f} + m_{2} * v_{2f} = 0.3 kg* (-0.86m/s) + 4.00 kg* v_{2f} (2)[/tex]
As p₀ = pf, from (1) and (2), we can solve for v2f, as follows:
[tex]v_{2f} = \frac{0.3kg* (1.3 + 0.86) m/s}{4.00kg} = \frac{0.3kg*2.16m/s}{4.00kg} = 0.16 m/s[/tex]
As the sign of v2f is positive, we conclude that it starts to move in the same direction that m1 was originally going (to the right), whilst m1 recoils, which means that after the collision, it bounces back from the larger mass.what is compressibility?
Hi! Compressibility is the capacity of something to be flattened or reduced in size by pressure. I hope this helps you! Good luck and have a great day. ❤️
34. If R = 4.0 kΩ, C = 3.0 mF, ε = 15 V, Q = 12 mC, and I = 2.0 mA, what is the potential difference Vb − Va?
Answer:the potential difference Vb − Va = -3 V
Explanation:
The potential difference Vb − Va s given by
Vb − Va= Capacitance / Charge + current x resistance - Electromotive force
Vb − Va=Q/C + I X R - E
V=Q / C=(12 x 10^-3) / (3 x 10^-3)
=0.012/0.003ss
=4V
alsoV= 1R = 2.0 x 10^-3 x 4 x 10^3
=0.002 X 4000
=8
Vb − Va= 4V + 8V- 15V
= -3V
Therefore, the potential difference Vb − Va = -3 V
last question, PLEASE help
An athlete wants your shoe company to make his/her new basketball shoe. He is known for his speed down the court and doesn't want his shoes to have any pattern on the outsole, the bottom of the shoe, because he thinks it would slow him down. How would you convince this athlete that a pattern is needed because friction is important in his game?
Answer:
my answer is in explanation below
Explanation:
You would convince him by saying the pattern adds elevation to the shoe therefore causing more friction.
(If it's wrong, I am sorry, I'm not in college)
(If it's right, can I get that brainliest?)
Which of the following is correct about a motion diagram?
a. A group of images for different objects found in one place at the same time
b. A series of images that show the position of a moving object at equal time intervals
c. A series of images that show the position of a moving object at unequal time intervals
d. Collection of photos taken by a person while moving around different places
Answer:
d. collection of photos taken by a person while moving around different places
Answer:
Moving object at equal time intervals
A 0.150-kg baseball traveling in a horizontal direction with a speed of 21.0 m/s hits a bat and is popped straight up with a speed of 15.0 m/s. If the bat was in contact with the ball for 51.0 ms, what was the average force of the bat on the ball
Answer:
17.65N
Explanation:
We know that the relationship for the impulse and momentum is given as
Ft=mv
given data
mass m=0.15kg
t= 51ms= 26.5/1000= 0.051seconds
velocity v1= 21m/s
Velocity v2=15m/s
substituting in the expression we have
Ft=m(v1-v2)
make F subject of formula we have
F=m(v1-v2) /t
F=0.15(21-15)/0.051
F=0.15*6/0.051
F=0.9/0.051
F=17.65N
This question involves the concepts of momentum and impulse.
The average force of the bat on ball is "17.65 N".
What is the average force?According to Impulse Equation the change in momentum is equal to the product of applied force and time of contact. Mathematically:
[tex]\Delta P=F\Delta t\\\\F=\frac{m\Delta v}{\Delta t}[/tex]
where,
F = Average force = ?m = mass of baseball = 0.15 kgΔv = change in speed = 21 m/s - 15 m/s = 6 m/sΔt = time of contact = 51 ms = 0.051 sTherefore,
[tex]F=\frac{(0.15\ kg)(6\ m/s)}{0.051\ s}[/tex]
F = 17.65 N
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A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. As the drawing shows, one person hits the water 5.00 m from the end of the slide in a time of 0.378 s after leaving the slide. Ignore friction and air resistance, find the height H in the drawing.
Answer:
9.63 m
Explanation:
We know that
Δx = vt.
Therefore, v = Δx / t
v = 5 / 0.378
v = 13.23 m/s
h = ½gt, so that
h = ½ * 9.8 * 0.378²
h = 4.9 * 0.1429
h = 0.7 m
Finally, using the law of conservation of energy, we know that
½mv² = mg(H - h)
½v² = g(H - h)
½v²/g = H - h, substituting the values and solving, we have
(½ * 13.23²) / 9.8 = H - 0.7
(½ * 175) / 9.8 = H - 0.7
87.5 / 9.8 = H - 0.7
8.93 = H - 0.7
H = 8.93 + 0.7
H = 9.63 m
Which type of power plant uses the energy of moving water in nature to
generate electricity?
A. Radiant power
B. Natural gas
C. Coal
D. Hydroelectric
Answer:
D
Explanation:
Answer:
Hydroelectric power plants
Explanation:
This is because hydropower is a form of renewable energy that uses water stored in dams, rivers to create electricity
The tendency of an object to resist change in its motion is known as
a. mass
b. inertia
c. force
d. balance
A quarterback applies a force 35 N for .28 seconds. What is the
impulse given to the ball?
A dairy farmer notices that a circular water trough near the barn has become rusty and now has a hole near the base. The hole is 0.29 m below the level of the water that is in the tank. If the top of the trough is open to the atmosphere, what is the speed of the water as it leaves the hole
Answer:
2.385 m/s
Explanation:
Here there is a transformation of potential energy of the water at the top to Kinectic energy at the hole
This can be expressed as
Potential energy= Kinect energy
mgh= 1/2m v²
Where m= mass of water
h= Lenght below the level of the water that is in the tank.
v= the speed of the water as it leaves the hole
g= acceleration due to gravity= 9.81m/s
Here there is 'm' at both sides which can cancelled beach other out.
Then we have
gh= ¹/₂v²
making v² subject of the formular
v²=gh/(¹/₂)
v=√(2gh)
V=√2g h
V= √(2×9.81×0.29)
V=2.385 m/s
Therefore, the speed of the water as it leaves the hole is 2.385 m/s
Which type of muscle helps the body move?
a
Cardiac
b
Smooth
c
Skeletal
d
None of the Above
Answer:
Cccccccc GIVE BRAINLYEST PLEASE
Answer: It´s c-skeletal
Explanation:
A piece of lead with a specific heat capacity of 126J/kg°C is given 5000J of energy to heat it from 20°C to 250°C. What was the mass of the piece of lead?
Answer:
m = 0.172 [kg]
Explanation:
To solve this problem we must use the expression of thermal energy that relates the heat to the mass, the specific heat, and the temperature differential.
Q = m*Cp*(T_f - T_i)
where:
m = mass [kg]
Cp = specif heat = 126 [J/kg*C]
Q = 5000 [J]
T_f = final temperature = 250 [C]
T_i = initial temperature = 20 [C]
m = Q/ (Cp*(T_f - T_i))
m = 5000/ (126*(250 - 20))
m = 0.172 [kg]
The mass of the moon is about 1/81 the mass of the earth and its radius is 1/4
that of the earth. The
acceleration due to gravity at the earth's surface is 9.8 m/s2. What is the acceleration due to gravity on the moon's surface?
9.8 m/s2
2.8 m/s2
1.9 m/s2
3.0 m/s2
What is the most logical explanation for why kilo is a more important prefix to remember than tera
What is an advantage of using nuclear power over natural gas to generate
electricity?
A. Nuclear power produces no waste.
B. Nuclear power does not contribute to climate change.
C. Nuclear power produces greenhouse gases.
D. Nuclear power is free.
Answer:b
Explanation:
I got the answer right
The main advantage of using nuclear power over natural gas to generate the electricity that nuclear power does not contribute to climate change, therefore the correct answer is option B.
What is a nuclear power plant ?It is a particular kind of power plant where electricity is produced using a nuclear reactor that may use nuclear fusion or nuclear fission .
When a radioactive substance undergoes a nuclear reaction, a tremendous quantity of energy is produced .
As given in the problem , we have to find out what is an advantage of using nuclear power over natural gas to generate electricity .
Option B is the proper response , since employing nuclear power instead of natural gas to produce electricity has the advantage of not causing the climate change .
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An asteroid floating in space attracts a skyscraper on Earth’s with gravitational force of 12
newtons from a given distance. If the distance between the two objects is reduced to half, what is the
changed force of attraction between them?
a. 3 Newtons
b. 24 Newtons
c. 48 Newtons
d. 144 Newtons
Answer:
Explanation:
it is C
Consider a car with a mass of 1250 kg moving at a constant speed of (22.0 A) m/s in a circular off ramp with a radius of (65.0 +B) m. What does the coefficient of static friction have to be for the car to just maintain its motion of ramp? Give your answer to 3 significant figures.
A=7
B=2
Answer:
Explanation:
weight of the car mg = m g
frictional force acting on the car = μ mg where μ is coefficient of friction
= m μ g
This force provides centripetal force , hence
m μ g = m v² / R
μ = v² / rg
= 22 x 22 / 65 x 9.8
= .759
a 45 kg girl is bouncing on a trampoline. during a certain interval after she leaves t he surface of the trampoline, her kinetic energy decreases to 220 J from 420 J. how high does she rise during this interval neglect air resistance
Answer: Height = 0.45m
Explanation:
According to the conservation of energy, When there are no interactions outside a system ie neglecting air resistance,The change in kinetic energy is equal to the change in potential energy in a closed system.
Therefore the kinetic energy decrease, the potential energy increases, and decreases for increase in Kinetic energy
an is represented by the formulae
ΔKinetic energy + ΔPotential energy = 0
ΔKinetic energy = -ΔPotential energy ( where negative sign represent decrease as kinetic energy increases.
ΔK.E=ΔP.E
420J- 220J= mgh
mass = 45kg
g =gravitational acceleration =9.8 m/s^2
200J = 45 X 9.8 X h
h = 200/45 x 9.8 /= 0.45m
Dangerous substances in an ecosystem that are harmful to its organisms are:
A sediments. B scavengers. C pollutants. D overpopulation.
Answer:
C pollutants
Explanation:
Pollutants are dangerous substances in the ecosystem that are harmful to its organisms. They are contaminants and effluents that leads to pollution in the environment.
There are different types of pollutants. Most of them are introduced into the environment by anthropogenic activities. Also, some natural events also introduce these pollutants into the environment. These materials pose huge health risks to living organisms. They can render the environment unfit for life.Answer:
c
Explanation: