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.
Learn more about the Impulse - momentum theorem here:
https://brainly.com/question/14121529
True or false: Electromagnetic radiation cannot travel through a vacuum.
A
True
B
False
Answer: that one is wrong i did it on flocabulary its false
Explanation:
Since electromagnetic radiations do not require a material medium for their propagation and can travel through vacuum, the statement that electromagnetic radiation cannot travel through a vacuum is False, option B.
What is an electromagnetic radiation?!An electromagnetic radiation is a radiation which results from.the interaction of the electric and magnetic fields.
Electromagnetic radiations do not require a material medium for their propagation, and hence can travel through vacuum.
An example of electromagnetic radiation is radio waves.
Hence, the statement that electromagnetic radiation cannot travel through a vacuum is False, option B.
Learn more about electromagnetic radiation at: https://brainly.com/question/7385447
I need help with this science work.
Answer:
ok what is it?
Explanation:
Answer:how can I help?
Explanation:
g First find the magnitude of the force F on a positive charge q in the case that the velocity v⃗ (of magnitude v) and the magnetic field B⃗ (of magnitude B) are perpendicular. Express your answer in terms of v, q, B, and other quantities given in the problem statement.
Answer:
The magnitude of the force F on a positive charge q in the case that the velocity v and the magnetic field B are perpendicular is [tex]F = q\cdot v\cdot B[/tex], measured in newtons.
Explanation:
From classical theory on Magnetism, the vectorial form of the magnetic force on a particle is given by:
[tex]\vec F = q\cdot \vec v \times \vec B[/tex] (Eq. 1)
Where:
[tex]\vec F[/tex] - Magnetic force, measured in newtons.
[tex]q[/tex] - Electric charge, measured in coulombs.
[tex]\vec v[/tex] - Velocity of the particle, measured in meters per second.
[tex]\vec B[/tex] - Magnetic field, measured in tesla.
By definition of cross product, we get that magnitude of magnetic force on a positive charge [tex]q[/tex] is:
[tex]F = q\cdot v\cdot B \cdot \sin \theta[/tex] (Eq. 2)
Where:
[tex]v[/tex] - Speed of the particle, measured in meters per second.
[tex]B[/tex] - Magnitude of the magnetic field, measured in tesla.
[tex]\theta[/tex] - Angle between the velocity of the particle and magnetic field, measured in sexagesimal degrees.
If velocity and magnetic field are perpendicular, then (Eq. 2) is reduced into this form: ([tex]\theta = 90^{\circ}[/tex])
[tex]F = q\cdot v\cdot B[/tex]
34)
Elements are composed of
A)
many type of atoms.
B)
the same type of atom.
C)
many different compounds.
D)
the same type of compound.
A vertical force is applied to a block of mass m that lies on a floor. What happens to the magnitude of the normal force on the block from the floor as magnitude F is increased from zero if force is (a) downward and (b) upward
Answer:
The answer is "[tex]\bold{+9.0 \ \frac{m}{s^2}}[/tex]"
Explanation:
Its rate (in respect of time) was its derivative of its specified function x, as well as the speed is the derivative of the speed. Therefore, in Newton's second legislation [tex]a=2c- 3(2.0)(2.0)t[/tex] is used:
[tex]F=(2.0 \ kg)a=40c -24t[/tex] (understood by SI units).
At [tex]t=3.0s, the \ F=-36 \ N[/tex] is told.
Therefore, the solution of c can be used to solve [tex]F=-36\ N=4.0c-24(3.0)[/tex]. This results in [tex]c=+9.0 \ \frac{m}{s^2}[/tex].
The graph below shows the average monthly temperatures at one location on earth over a three-year period.
Which was the highest record temperature?
A:25°F
B:30°F
C:70°F
D:80°F
A 27 kg child slides down a playground slide at a constant speed. The slide has a height of 4.0 m and is 7.0 m long. Part A Using the law of conservation of energy, find the magnitude of the kinetic friction force acting on the child.
Answer:
Explanation:
The forces acting along the playground are:
Kinetic friction force Fk and
Wsintheta (weight along the plane)
Since sintheta = opp/hyp
Sintheta = 4/7
Taking the sum of forces along the playground
\sumFx = max
Wsintheta - Fk = max
Since the speed is constant, ax = 0, the equation becomes
Wsin theta - Fk = 0
Fk = Wsintheta
From the question
W = mg
W = 27×9.81
W = 264.87N
Fk = 264.87 × 4/7
Fk = 1059.48/7
Fk = 151.35N
Hence the magnitude of the kinetic friction force acting on the child is 151.35N
Help please, thank you.
Answer:
a. skin and mucous membranes
hope this helps you
❤❤❤❤❤❤❤
itz Indian heart
At an instant when a soccer ball is in contact with the foot of the player kicking it, the horizontal or x component of the ball's acceleration is 810 m/s2 and the vertical or y component of its acceleration is 910 m/s2. The ball's mass is 0.40 kg. What is the magnitude of the net force acting on the soccer ball at this instant?
Answer:
487.13N
Explanation:
According to Newton's second law;
F = ma
F is the net force
m is the mass of the object
a is the acceleration
First we need to find the resultant of the acceleration
a =√810²+910²
a = √656,100+828,100
a =√1,484,200
a = 1218.28m/s²
Ball mass = 0.40kg
Next is to get the magnitude of the net force
F = 0.4 × 1218.28
F = 487.31N
Hence the magnitude of the net force acting on the soccer ball at this instant is 487.13N
Which statement is true regarding eligibility for becoming pharaoh?
People believed that the gods selected the pharaoh.
A woman could be selected as pharaoh.
The pharaoh’s title had to pass to a royal son.
The title of pharaoh was passed to a military commander.
Answer:
A woman could be selected as pharaoh.
Explanation:
Most pharaohs were men, but there were some exceptions, such as queen Hatshepsut.
Answer:
A
Explanation:
Starting at rest at the top of a ski slope, you travel down the hill reaching a velocity of 20 m/s after 13 seconds. Once you reach the bottom of the hill you travel at a constant velocity of 20 m/s for the next 7 seconds. Sketch a quantitative (with numbers) position vs time, velocity vs time, and acceleration vs time graph of the skier's motion. PLEASE SHOW WORK AND I WILL MARK YOU THE BRAINLIEST
Answer:
The graphs are attached
Explanation:
We are told that he started at rest at rest and travelled down the hill reaching a velocity of 20 m/s after 13 seconds.
Acceleration is gotten from;
v = u + at
a = (v - u)/t
a = (20 - 0)/13
a = 20/13 m/s² or 1.54 m/s²
Distance in this period is gotten from;
v² = u² + 2as
s = (v² - u²)/2s
s = (20² - 0²)/(2 × 20/13)
s = 400/(40/13)
s = 130 m
We are told that after reaching the bottom of the hill, he travelled at a constant velocity of 20 m/s for the next 7 seconds.
At constant velocity, acceleration is 0.
Thus,distance in this period is;
s1 = vt = 20 × 7 = 140 m
I've attached the graphs
A block of mass, m, is pushed up against a spring, compressing it a distance x, and is then released. The spring projects the block along a frictionless horizontal surface, giving the block a speed v. The same spring projects a second block of mass 4m, giving it a speed 3v. What distance was the spring compressed in the second case?
Answer:
x₂ =6*x₁
Explanation:
In absence of friction, along a horizontal surface, all the elastic potential energy stored in the spring becomes kinetic energy, so we can write the following expression for the first case:[tex]\frac{1}{2}*k*x_{1}^{2} = \frac{1}{2}*m*v^{2} (1)[/tex]
In the second case, we have a mass 4m, which reaches to a speed 3v due to the energy stored in the spring.We can write the same equation than for (1) replacing m by 4m and v by 3v, as follows:[tex]\frac{1}{2}*k*x_{2} ^{2} = \frac{1}{2} * 4*m*(3*v)^{2} =\frac{1}{2}*4*m*9*v^{2} = \frac{1}{2}* 36 m*v^{2} (2)[/tex]
Dividing (2) over (1) on both sides, and rearranging terms, we get:[tex]x_{2} ^{2} = 36 * x_{1} ^{2}[/tex]
Taking square roots on both sides, we get:x₂ = 6*x₁So, in this case, the spring was compressed six times the distance that was compressed in the first case.If I am driving down the highway going north at 50 miles per hour, and another car is driving south at 75 miles per hour. How fast is the car coming toward me?
Answer:
V₂₋₁ = 50 [miles/h]
Explanation:
In order to solve this problem we must use the concept of relative velocities.
Where:
V1 = 50 [miles/h]
V2 = 75 [miles/h]
V₂₋₁ = Relative velocity among the two cars.
V₂₋₁ = 75 - 25 = 50 [miles/h]
That is, the car that goes at 50 miles per hour, sees how the second car approaches at 50 miles per hour.
frictionless floor by force . What total mass is accelerated to the right by (a) force , (b) cord 3, and (c) cord 1
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]F = 20 \ kg[/tex]
b
[tex]F = 18 \ kg[/tex]
c
[tex]F = 10 \ kg[/tex]
Explanation:
From the diagram we are told that
The first mass is [tex]m_1 = 10 \ kg[/tex]
The second mass is [tex]m_2 = 3 \ kg[/tex]
The third mass is [tex]m_3 = 5 \ kg[/tex]
The fourth mass is [tex]m_4 = 2 \ kg[/tex]
Gnerally the total mass accelerated by right force is mathematically evaluated as
[tex]F = m_1 + m_2 +m_3 + m_4[/tex]
[tex]F = 10 + 3 +5 + 2[/tex]
[tex]F = 20 \ kg[/tex]
Gnerally the total mass accelerated by cord 3 is mathematically evaluated as
[tex]F = m_1 + m_2 +m_3[/tex]
[tex]F = 10 + 3 +5[/tex]
[tex]F = 18 \ kg[/tex]
Gnerally the total mass accelerated by cord 3 is mathematically evaluated as
[tex]F = m_1[/tex]
[tex]F = 10 \ kg[/tex]
A ballistic pendulum is a device often used for measuring the velocity of a projectile. Assume to projectile fully embeds itself in the pendulum. If the projectile mass is 25 grams, the pendulum mass is 2.5 kg, and the vertical displacement of the pendulum after the collision is 12 cm, find the velocity of the projectile just prior to the collision.
Answer:
Explanation:
pendulum rises upto height of .12 m
velocity of projectile and pendulum after collision = √ ( 2gH )
= √ ( 2 x 9.8 x .12 )
= 1.53 m / s
Applying conservation of momentum during collision
mv = (m + M ) X V
m and M are mass of projectile and pendulum and v and V are velocity of projectile before and after collision .
.025 x v = ( 2.5 + .025 ) x 1.53
v = 154.5 m /s
2. A 60 kg crate is being lowered from a loading dock to the floor of a warehouse by a thick rope. The rope exerts a constant upward force of 500 N as it is descending to the floor. a) Draw a free-body diagram for the crate. b) Calculate the acceleration (if any) of the crate.
Answer:
Acceleration =8.33m/s^2Explanation:
Step one:
Kindly find attached a sketch of the free body diagram
Given data
mass m=60kg
Force F=500N
a=?
Step two:
From Newton's second law "An object’s acceleration equals the vector sum of the forces acting on it, divided by it mass".
mathematically
F=ma
Step three:
Substituting we have
500=60*a
a=500/60
a=8.33m/s^2
An athlete at the gym holds a 3.5kg steel ball in his hand. His arm is 70cm long and has a mass of 4.0kg.What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor?What is the magnitude of the torque about his shoulder if he holds his arm straight, but 55 degrees below horizontal?
Answer:
A. 38 nm
B. 22 nm
Explanation:
This question has given us m, mass of the ball as 3.5kg
M = The mass of arms = 4kg
R = length of arms as 70cm
70 cm = 0.7m
0.7m/2 = 0.35m
rmg + Rmg
g = 9.8m/s
= 0.7(3.5)(9.8)+(0.35)(4)(9.8)
= 24.01 + 13.72
= 37.72
~38
B.
55 degrees below horizontal
Cos 55⁰ = 0.5736
= 0.7(3.5)(9.8)(0.5736)+(0.35)(4)(9.8)(0.5736)
= 13.772136+7.869792
= 21.642
~22Nm
The magnitude of the torque about his shoulder, if he held his arm straight is 21.64Nm
The formula for calculating the magnitude of the torque about his arm is expressed as:
[tex]\tau = (F_1r + F_2R) cos\theta[/tex]
F1 and F2 are the forces
[tex]\theta[/tex] is the given angle
F1 = 3.5 * 9.8 = 34.3 N
r = 70cm = 0.7m
F2 = 4.0 * 9.8 = 39.2N
R = 0.7/2 = 0.35m
Substitute the given values into the formula:
[tex]\tau = (34.3(0.7) + (39.2)(0.35)) cos55^0\\\tau = (24.01+13.72)cos55\\\tau = 37.73cos55\\\tau=21.64Nm[/tex]
Hence the magnitude of the torque about his shoulder, if he held his arm straight is 21.64Nm
Learn more here: https://brainly.com/question/21603137
A car going around a curve of radius R at a speed V experiences a centripetal acceleration ac. What should be the radius of the path that leads to same centripetal acceleration while it travels at speed 2V
Answer:
4R
Explanation:
The centripetal acceleration a is given by a = v²/r where V = speed of object and r = radius of circle. Now, it is given that V = speed of car = V and radius of circle = R, so its centripetal acceleration a = V²/R.
Now, if the speed is doubled to V' = 2V with the same centripetal acceleration, we find the new radius of the curve R' from
a = V'²/R'
R' = V'²/a
R' = (2V)² ÷ V²/R
R' = 4V²R/V²
R' = 4R
So, the new radius is 4 times the initial radius.
A ball rolls at a speed of 4 m/s off a flat surface which is a height 1.3m above the ground. how long (in seconds) is the ball in the air before it hits the ground
Determine the xx component of force at point CC in the pipe assembly. Neglect the weight of the pipe. Take F1F1
Answer:
Hello your question has some missing part and diagram attached below is the missing part and diagram
Determine the x component of force at point C in the pipe assembly. Neglect the weight of the pipe. Take F1 = { 348 i - 427 j} lb and F2 = {-307 j + 122 k} lb.
Determine the y component of the force at point C in the pipe assembly.
Determine the x component of the force at point C in the pipe assembly.
Determine the x component of the moment at point C in the pipe assembly.
Determine the y component of the moment at point C in the pipe assembly.
Determine the z component of the moment at point C in the pipe assembly.
answer: attached below is the detailed solution
Explanation:
Attached below is the detailed solution
The velocity of an object as a function of time is given by v = (12.5t –7.2t2) + (4.3t3). What is its acceleration as a function of time?
Answer:
a(t) = 12.5 -10.4t + 12.9t²Explanation:
The equation is not well written:
let the velocity of an object as a function of time be given by
v = (12.5t - 7.2t²) + (4.3t³)
v(t) = 12.5t - 7.2t² + 4.3t³
acceleration is the change in velocity with respect to time:
a(t) = d(v(t))/dt
a(t) = 12.5 - 2(7.2)t^2-1 + 3(4.3)t^3-1
a(t) = 12.5 -10.4t + 12.9t²
Hence the acceleration as a function of time t is expressed as;
a(t) = 12.5 -10.4t + 12.9t²
Very Urgent!!!!!!! Please!!!!!!!!
A 35kg child standing on ice (frictionless) pushes a 7kg bowling ball with a force of 100N. What is the force acting on the 35kg child? What is the acceleration of the 35kg child?
Please real answers!! Gives many points!!!!!!
Answer:
a = 2.85714 m/s^2
Explanation:
Fnet=ma
eaqual and oppisate forces means that the 100N pushed on the ball comes back to the child.
100 = 35*a
a= 100/35
A 0.1-kg meter stick is supported at both ends by strings, and there is a 200-g mass attached to it at the 70 cm mark. What is the force exerted by the string connected at the 0 cm mark, if the system is in static equilibrium
Answer:
The force exerted by the string connected at the 0 cm mark is 1.078 N
Explanation:
Given;
mass of the meter stick, m = 0.1 kg
weight of the meter stick, = 0.1 kg x 9.8 m/s² = 0.98 N
the weight attached at 70 cm mark = 0.2 kg x 9.8 m/s² = 1.96 N
0cm--------------------------50cm-----------70cm-------------100cm
↑ ↓ ↓ ↑
F₁ 0.98N 1.96N F₂
Take the moment about F₂: clockwise moment = anticlockwise moment
F₁(100 - 0) = 0.98(100 -50) + 1.96(100 -70)
100F₁ = 49 + 58.8
100F₁ = 107.8
F₁ = 107.8 / 100
F₁ = 1.078 N
Therefore, the force exerted by the string connected at the 0 cm mark is 1.078 N
Why is force a vector
Answer:
A force vector is a representation of a force that has both magnitude and direction.
A 60 kg cyclist approaches the bottom of a gradual hill at a speed of 11 m/s. The hill is 5.0 m high, and the cyclist estimates that she is going fast enough to coast up and over it without peddling. Using conservation of energy, find the speed of the cyclist when she reaches the top of the hill (ignoring air resistance and friction). g
Answer:
9.91m/s
Explanation:
According to conservation of energy, the potential energy at the top of the hill is equal to the kinetic energy along the hill.
PE = KE
mgh = 1/2mv²
m is the mass of the cyclist
g is the acceleration due to gravity
h is the height of the hill
v is the speed of the cyclist
From the formula:
gh = v²/2
Given
g = 9.81m/s²
h = 5.0m
Substitute into the formula:
9.81(5.0) = v²/2
9.81*5*2 = v²
98.1 = v²
v = √98.1
v = 9.91m/s
Hence the speed of the cyclist when she reaches the top of the hill is 9.91m/s
A 3.3 kg block of copper at a temperature of 74°C is dropped into a bucket containing a mixture of ice and water whose total mass is 1.2 kg. When thermal equilibrium is reached the temperature of the water is 8°C. How much ice was in the bucket before the copper block was placed in it? (Neglect the heat capacity of the bucket.
Answer:
0.228 kilograms of ice were in the bucket before the copper block was placed in it.
Explanation:
In this case, we assume that water inside the bucket was in the form of ice at a temperature of 0 ºC and an atmospheric pressure of 101.325 kilopascals, It is also known that block of copper is cooled down whereas ice is melted and heated up until thermal equilibrium is reached. If ice-block system is an isolated system, then First Law of Thermodynamics depicts the following model:
[tex]-Q_{b} +Q_{l,w}+Q_{s, w} = 0[/tex]
[tex]Q_{b} = Q_{l,w}+Q_{s,w}[/tex] (Eq. 1)
Where:
[tex]Q_{b}[/tex] - Heat released by the block of copper, measured in kilojoules.
[tex]Q_{l,w}[/tex] - Latent heat received by water, measured in kilojoules.
[tex]Q_{s,w}[/tex] - Sensible heat received by water, measured in kilojoules.
By definitions of sensible and latent heat, we expand the equation as follows:
[tex]m_{b}\cdot c_{b}\cdot (T_{b,o}-T) = m_{w}\cdot [L_{f}+c_{w}\cdot (T-T_{w,o})][/tex] (Eq. 2)
Where:
[tex]m_{b}[/tex] - Mass of the block of copper, measured in kilograms.
[tex]c_{b}[/tex] - Specific heat of copper, measured in kilojoules per kilogram-degree Celsius.
[tex]T_{b,o}[/tex] - Initial temperature of block of copper, measured in degrees Celsius.
[tex]m_{w}[/tex] - Mass of water, measured in kilograms.
[tex]L_{f}[/tex] - Latent heat of fusion of water, measured in kilojoules per kilogram.
[tex]T_{w,o}[/tex] - Initial temperature of water, measured in degrees Celsius.
[tex]T[/tex] - Final temperature of water-block system, measured in degrees Celsius.
[tex]c_{w}[/tex] - Specific heat of water, measured in kilojoules per kilogram-degree Celsius
And we clear the mass of water of the system:
[tex]m_{w} = \frac{m_{b}\cdot c_{b}\cdot (T_{b,o}-T)}{L_{f}+c_{w}\cdot (T-T_{w,o})}[/tex]
If we know that [tex]m_{b} = 3.3\,kg[/tex], [tex]c_{b} = 0.385\,\frac{kJ}{kg\cdot ^{\circ}C}[/tex], [tex]T_{b,o} = 74\,^{\circ}C[/tex], [tex]T = 8\,^{\circ}C[/tex], [tex]L_{f} = 334\,\frac{kJ}{kg}[/tex], [tex]c_{w} = 4.186\,\frac{kJ}{kg\cdot ^{\circ}C}[/tex] and [tex]T_{w,o} = 0\,^{\circ}C[/tex], then the mass of the ice inside the bucket is:
[tex]m_{w} = \frac{(3.3\,kg)\cdot \left(0.385\,\frac{kJ}{kg\cdot ^{\circ}C} \right)\cdot (74\,^{\circ}C-8\,^{\circ}C)}{334\,\frac{kJ}{kg}+\left(4.186\,\frac{kJ}{kg\cdot ^{\circ}C} \right)\cdot (8\,^{\circ}C-0\,^{\circ}C) }[/tex]
[tex]m_{w} = 0.228\,kg[/tex]
0.228 kilograms of ice were in the bucket before the copper block was placed in it.
A bullet is fired directly upward with a speed of 330 m/s. How far above the ground will the bullet be in 65 s
Answer:
d = 21450 m
Explanation:
Given that,
The speed of a bullet, v = 330 m/s
We need to find the distance covered by the bullet in 65 s.
We know that,
Speed of an object = distance covered divided by time taken.
It means,
d = vt
d = 330 m/s × 65 s
d = 21450 m
So, the bullet will cover a distance of 21450 m.
Question 3: In one half hour, a car traveled 25.2 km. Its average speed is
50.4 km/min
12.6 km/hr
14 m/s
none of them
Answer:
12.6 km/hr
explanation:
25.2 x [tex]\frac{1}{2}[/tex]
Which is the dependent variable and what is the trend in the graph?
Experimental Solubility Data for a Sugar
140
130
120
110
Solubility (g/100g H,0)
100
90
80
0
10
50
60
20 30 40
Temperature (*C)
Solubility; decreasing
Solubility; increasing
Temperature; decreasing
Temperature; increasing
Answer: Solubility; decreasing
Explanation:
A person throws a shot put from height of 5 feet with an initial vertical velocity of 40 feet per second. Use the vertical motion model, h = -16t^2 + vt + s, where v is the initial velocity in feet per second and s is the initial height in feet, to calculate the amount of time the shot put is in the air before it hits the ground.
Answer:2.6
Explanation: