In riot control, the riot squad uses a water hose that shoots water at a rate of 5 m/s and volume of 30 L/s. What is the average force exerted on a person assuming that the water splashes sideways in all directions? The density of water is 1000 kg/m3.

Answer:

Explanation:

A 52.0 kg skateboarder wants to just make it to the upper edge of a "quarter pipe," a track that is one-quarter of a circle with a radius of 2.60 m .

What speed does he need at the bottom

Let the lowest point of the circle = (0,0)

Center of circle = (0, 2.6)

The height of the circle = 2 * radius = 2 * 2.60 = 5.20 m

Highest point = (0, 5.2)

As the skateboarder moves around ¼ of a vertical circle, the skateboarder moves from the lowest position to a position that is ½ the way up to the highest position. This is the point that is 2.6 meters directly to the right of left of the center = (2.6, 2.6)

As the skateboarder has moved 2.6 m upward, the potential energy increase = m * g * ∆h = 52.0 * 9.8 * 2.6

During this same time, the kinetic energy decreased from the maximum to 0. The decrease of KE = the increase of PE

½ * m * ∆v^2 = m * g * ∆h

½ * 52 * ∆v^2 = 52.0 * 9.8 * 2.6

½ * ∆v^2 = 9.8 * 2.6

∆v = (2 * 9.8 * 2.6)^0.5 = 7.14 m/s

The velocity at highest point, (2.6,2.6) is 0 m/s

So the velocity at the lowest point must be 7.14 m/s

Let’s see if the skateboarder has enough KE to move upward 2.6 m.

KE at bottom = ½ * 52 * 7.14^2 = 1325.5

PE = 52 * 9.8 * h

52 * 9.8 * h = 1325.5

h = 2.6 m

The answer is OK

The speed needed by the skateboarder at the bottom is 7.8 m/s.

The given parameters;

Apply the principle of conservation of mechanical energy to determine the sped of the skateboarder at the bottom;

[tex]P.E_{top} = K.E_{bottom}\\\\mgh = \frac{1}{2}mv^2\\\\gh = \frac{1}{2} v^2\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2\times 9.8 \times 3.1} \\\\v = 7.8 \ m/s[/tex]

Thus, the speed needed by the skateboarder at the bottom is 7.8 m/s.

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Answer:

The answer is "[tex]\bold{MB= \frac{L^2 \ WB}{12}}[/tex]"

Explanation:

please find the complete question in the attached file:

From A to B  

[tex]reqAB = \int^{\frac{1}{2}}_{0} \frac{WB}{\frac{L}{2}} x \ dx = \frac{L\ WB}{4}[/tex]

From symmetry [tex]Ay = Cy = FeqAB[/tex]  

The FeqAB position is found on a table of common centres:

[tex]\frac{2}{3}( \frac{L}{2})[/tex]

[tex]\to \Sigma_{NB=0} FX = NB=0\\\\\to \Sigma_{VB=0} Fy = \frac{LWB}{4} - \frac{LWB}{4} -VB =0 \\\\\to \Sigma M_{.CCW \ A} = \frac{L}{3} - \frac{LWB}{4} - \frac{L}{2}VB+MB =0 \\\\MB= \frac{L^2 \ WB}{12}[/tex]

Answer:

2.54*10^8 m/s

Explanation:

Given that

Speed of light in space, c = 299792458 m/s

Wavelength of the beam of light, λ = 621 nm or 621*10^-9 m

Index of refraction of the liquid, n = 1.18

Speed of light in the liquid, v = ?

The index of refraction of an optical material, n is given by the following formula.

n = c/v, and if we substitute directly, we have

1.18 = 299792458 / v, if we make v subject of formula, we have

v = 299792458 / 1.18

v = 2.54*10^8 m/s

Answer:

The fundamental frequency is 10 Hz

Explanation:

The frequency of the fundamental mode of the string is given by;

[tex]f_o =\frac{1}{2l}\sqrt{\frac{T}{\mu }[/tex]

where;

L is length of the string = 2.5 m

T is tension of the string, = 10 N

μ is mass per unit length = m / L = 0.01 / 2.5 = 0.004 kg/m

Substitute the given values and solve for the frequency;

[tex]f_o =\frac{1}{2l}\sqrt{\frac{T}{\mu }}\\\\ f_o =\frac{1}{2*2.5}\sqrt{\frac{10}{0.004}}\\\\ f_o = 0.2\sqrt{2500}\\\\ f_o = 0.2 *50\\\\f_o = 10 \ Hz[/tex]

Therefore, the fundamental frequency is 10 Hz

Answer:

answer is C

Explanation:

Answer: Monetary policy used during times of recession

Explanation:

Answer:

The question is incomplete. However the complete question is :

A Rail Gun uses electromagnetic forces to accelerate a projectile to very high velocities. The basic mechanism of acceleration is relatively simple and can be illustrated in the following example. A metal rod of mass m and electrical resistance R rests on parallel horizontal rails (that have negligible electric resistance), which are a distance L apart. The rails are also connected to a voltage source V, so a current loop is formed.

The vertical magnetic field, initially zero, is slowly increased. When the field strength reaches the value [tex]$B_0$[/tex], the rod, which was initially at rest, begins to move. Assume that the rod has a slightly flattened bottom so that it slides instead of rolling. Use g for the magnitude of the acceleration due to gravity.

Find μs, the coefficient of static friction between the rod and the rails.

Express the coefficient of static friction in terms of variables given in the introduction.

So the answer is : [tex]$\mu_s=\frac{\Delta V LB_0}{Rmg}$[/tex]

Explanation:

It is given : B = [tex]$B_0$[/tex]

Therefore, using force on the current carrying wire is:

[tex]$\mu_s mg = B_0IL$[/tex]

[tex]$\mu_s= \frac{ILB_0}{mg}$[/tex]

Therefore,

[tex]$\mu_s=\frac{\Delta V LB_0}{Rmg}$[/tex]

The coefficient of static friction in terms of variables will be [tex]\rm \mu_s =\frac{ \triangle VLB_0}{Rmg} \\\\[/tex].

The ratio of the greatest static friction force (F) between the surfaces in contact before movement begins to the normal (N) force is the coefficient of static friction.

The given data in the problem is;

[tex]\rm B=B_0[/tex]

The force on a current-carrying wire is balanced by the friction force.

The force on a current-carrying wire is given by;

[tex]\rm \mu_s mg= B_0IL \\\\ \rm \mu_s=\frac{ILB_0}{mg}\\\\ \rm \mu_s= \frac{\triangle VLB_0}{Rmg}[/tex]

Hence the coefficient of static friction in terms of variables will be [tex]\rm \mu_s =\frac{ \triangle VLB_0}{Rmg} \\\\[/tex].

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Answer:

sciencephysicsphysics questions and answersA 12-kg Block Is Pressed Against A Spring (spring Constant 870 N/m ), Compressing It Some Distance. ...

Question: A 12-kg Block Is Pressed Against A Spring (spring Constant 870 N/m ), Compressing It Some Distance. The Block Is Released From Rest And Slides Across A Track As Shown In (Figure 1). While Most Of The Track Is Frictionless, There Is A 55-cm Section Of Track That Has A Coefficient Of Friction With The Block Of 0.4. A Bit Further On, The Track Ascends ...

A 12-kg block is pressed against a spring (spring constant 870 N/m ), compressing it some distance. The block is released from rest and slides across a track as shown in (Figure 1). While most of the track is frictionless, there is a 55-cm section of track that has a coefficient of friction with the block of 0.4. A bit further on, the track ascends into a hill that is 40-cm tall.

What is the minimum compression of the spring necessary for the block to slide past the section with friction?

What is the minimum compression of the spring necessary for the block to make it to the top of the hill on the other side? 40cm 12kg Ww.w.wwwww 55cm

Explanation:

Answer:

Explanation:

Let force P = 193 pounds

second force = Q

Angle that resultant makes with force P is θ

Tanθ = Q sin86.27 / (P + Q cos86.27 )

Tan 31.4 = Q sin86.27 / (193 + Q cos86.27 )

.61 = .99 Q / (193 + .065Q )

117.73 + .04 Q = .99Q

117.73 = .95 Q

Q = 123.93  pounds .

Answer:

The final speed of the electron is 6.626 x 10⁷ m/s

Explanation:

force applied to the electron, f = 4 x 10⁻¹⁵ N

distance traveled by the electron, d = 50 cm = 0.5 m

The work done on the electron;

W = F x d

W = (4 x 10⁻¹⁵ N) x (0.5 m)

W = 2 x 10⁻¹⁵ J

The kinetic energy of the electron is given by;

[tex]K.E = \frac{1}{2}m_ev^2[/tex]

Apply work - energy theorem;

K.E = W

[tex]\frac{1}{2}m_ev^2 = 2 *10^{-15}\\\\ v^2 = \frac{2( 2 *10^{-15})}{m_e}\\\\v= \sqrt{\frac{2( 2 *10^{-15})}{m_e}}\\\\ v = \sqrt{\frac{2( 2 *10^{-15})}{9.11*10^{-31}}}\\\\v = 6.626*10^7 \ m/s[/tex]

Therefore, the final speed of the electron is 6.626 x 10⁷ m/s

Answer:

s₂ = 0.475 m = 47.5 cm

Explanation:

The arc length and the angle of rotation are related through the formula:

s = rθ

where,

s = arc length

r = radius of curvature

θ = angle of rotation

First, we consider the arc length covered by the point of insertion of extensor muscles.

s₁ = r₁θ

where,

s₁ = arc length covered by insertion of extensor muscle = 5 cm

r₁ = length of insertion from knee = 4 cm

θ = Angle of Rotation = ?

Therefore,

5 cm = (4 cm)(θ)

θ = (5 cm)/(4 cm)

θ =  1.25 rad

Now, we consider the arc length covered by the foot.

s₂ = r₂θ

where,

s₂ = arc length covered by the foot = ?

r₂ = distance from knee to foot = 38 cm = 0.38 m

The angle of rotation will be the same for the foot as the insertion.

Therefore,

s₂ = (0.38 m)(1.25 rad)

s₂ = 0.475 m = 47.5 cm

Answer:

0.176

Explanation:

(5.27)/(3.05*9.8)

It is indeed 0.176 for the answer.

Answer:

It is increasing .60 m/s

Explanation:

Answer:

These two objects hit the ground at the same time. But their impact velocity will be different

because they are dropped from different heights.

The one that is dropped from the building will have more impact when it hits the ground because gravity causes this onject to fall toward the ground at a faster and faster velocity the longer the object falls. In fact, its velocity increases by 9.8 m/s2.

Explanation:

Initial velocity of the object u=0

Final velocity of the object when it reaches the ground v=A m/s

Using v^2−u^2=2gS where g=10m/s^2

A^2 −0^2=2(10)H

A^2=20*H H- heigh from the table

Object falling from the building will have the same initial velocity but different heigh.

T>H

so

B^2=20*T

20*T>20*H

so B^2>A^2

it equal to B>A so object falling from the building will have more impact than falling from the table.

Answer:

C - 98j

Explanation:

Answer:

c- 98j

hope it helped

Answer:

The object on the moon has greater mass

Explanation:

Given;

weight of the object on Earth, W₁ = 1 N

wight of the object on Moon, W₂ = 1 N

gravity on Earth, g₁ = 9.8 m/s²

gravity on the Moon, g₂ = g₁/6 = 1.633 m/s²

Apply Newton's second law of motion;

Weight of the object on Earth is given by;

W₁ = m₁g₁

m₁ = W₁/g₁

m₁ = 1 / 9.8

m₁ = 0.102 kg

Weight of the object on Moon is given by;

W₂ = m₂g₂

m₂ = W₂ / g₂

m₂ = 1 / 1.633

m₂ = 0.612 kg

Therefore, the object on the moon has greater mass.

Answer:

0.04558

Explanation:

Luminosity is defined as the total amount of energy a star can produce in a second. Luminosity is given by the formula

σ * A * T⁴

where;

σ = Stefan-Boltzmann constant

= 5.670367 * 10⁻⁸

A= Surface Area

=4πr²

T= Temperature in Kelvin

In the question,

Temperature= 4000K

Radius= 2.0 A.U

Substituting the variables into the equation,

0.000000056704 * 4(3.14)(2.0²) * 4000⁴

= 0.000000056704 * 50.24 * 16000

= 0.04558

Answer:

Explanation:

Mars does not appear to have any water, while Earth is mostly water. Mars also has yellow-brown sky (pink-brown at sunrise and sunset), which suggests that it has a different atmospheric makeup.

Answer:

Mars does not appear to have any water, while Earth is mostly water. Mars also has yellow-brown sky (pink-brown at sunrise and sunset), which suggests that it has a different atmospheric makeup.

Explanation:

Angular velocity of the coil :

[tex]\omega=\dfrac{2\pi n}{60}[/tex]

[tex]\omega=\dfrac{2\times \pi\times 3500}{60}\\\\\omega =366.52\ rad/s[/tex]

Now,

Peak voltage is given by :

[tex]V=NAB\omega\\\\V= 200\times \dfrac{\pi (0.1)^2}{4}\times 0.7\times 366.52\\\\V=403.01\ V[/tex]

Hence, this is the required solution.

Answer:

a) They both hit the ground at the same time

b) Watermelon will hit first, since its speed is faster than the pumpkin speed

c) h = 29,6 m

d)t = 3,02 sec

Explanation:

Equations for fall free movement are:

vf = v₀ + g*t       when   v₀ = 0   (dropped case)  vf = g*t

h = v₀*t + 1/2*g*t²

a) For both ( watermelon and pumpkin) the equation of speed is the same:

vf = g*t²    Both will have the same speed second through second

They both hit the ground at the same time

b) Now is watermelon is thrown with v₀ = 10 m/s

Watermelon will hit first since its speed is faster than the pumpkin speed

vf(watermelon) =  10 + g*t

vf₂ (pumpkin)   =  g*t

c) h = v₀*t + (1/2)*g*t²

h = (10)*1 + (1/2)*9,8*1

h = 10 + 19,6

h = 29,6 m

d)  h = g*t

t = 29,6/9,8

t = 3,02 sec

Answer:

[tex]P=V^2/R[/tex]

[tex]R=V^2/P[/tex]

Upon substituting the values

a) R= 75*75/7800 = 0.72ohm

   Also

   R=ρ*L/A

   L=R*A/ρ

   L=[tex]\frac{0.72*3.73*10^-6 }{5*10^-7}[/tex] = 5.44m

b) when the voltage 107 is used

    R=107*107/7800 = 1.468OHM

     L=[tex]\frac{1.468*3.73*10^-6 }{5*10^-7}[/tex] =11.1m

Answer:

That the universe is at least 13.2 billion years old

Explanation:

The most distant galaxy we have observed is more than 13.2 billion light-years away. this evidence indicates. That the universe is at least 13.2 billion years old

The universe is the sum total of all existent matter and space. The universe is estimated to be at least 10 billion light-years across and contains an enormous number of galaxies;

It has been expanding since the Big Bang some 13 billion years ago.

The most distant galaxy we have observed is more than 13.2 billion light-years away. this evidence indicates. That the universe is at least 13.2 billion years old

Hence option c is correct.

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Answer:

Locating misplaced electrical equipment.

Explanation:

Ultrasound is primarily used by doctors to locate fractured bones or damaged organs, so I assume it can be used to locate other things too.

Answer:

900J

Explanation:

The Work done can be calculated using

Work done = force × distance

According to this question, force = 300N, distance = 3m

Work done = 300 × 3

Work done = 900J

Gray exerted 900J of work

Answer:

mk7000000.0000000000

Answer:

v = 38.34 m

Explanation:

Given that,

Mass of a skier, m = 75 kg

It falls down a 75.0 m high slope without friction, h = 75 m

We need to find the velocity of the skier at the bottom of the slope. It is based on the concept of conservation of energy. So, v is the required velocity. So,

[tex]mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 75} \\\\v=38.34\ m/s[/tex]

So, the velocity of the skier at the bottom of the slope is 38.34 m.

Answer:

the vibration of the glass tube creating sound waves within itself.  reason is the rising pitch is that the liquid rise shortens the length of the vibrating area within the tube.

Explanation:

Answer:

[tex]B=3.29\ \mu T[/tex]

Explanation:

Given that,

Length of a wire is 3 m

Current in the wore, I = 10 A

The wire formed a semicircle.

We need to find the magnitude of the magnetic field (in micro-tesla) at the center of the circle along which the wire is placed.

The magnetic field at the center of semicircle is given by :

[tex]B=\dfrac{\mu_o I}{4R}[/tex]

R is radius of semicircle,

[tex]R=\dfrac{3}{\pi}=0.954\ m[/tex]

So,

[tex]B=\dfrac{4\pi \times 10^{-7}\times 10I}{4(0.954)}\\\\B=3.29\times 10^{-6}\ T\\\\B=3.29\ \mu T[/tex]

So, the magnitude of the magnetic field is [tex]3.29\ \mu T[/tex].

Answer:

D. V/3.

Explanation:

V = rw

At point a

Linear velocity

= Va = rw

At point b

We have,

V = 3rw

So linear velocity of a

= V/Va = 3rw/re

I cross multiples and after which I got Va = v/3

Please view attachment

Answer:

Explanation:

The relation between linear and angular velocity,

[tex]V = rw[/tex]

Now, for point A the linear velocity is

[tex]V_A = rw[/tex]

And for point B

[tex]V = 3rw[/tex]

Therefore, the linear velocity of the point A

[tex]\frac{V}{V_A} = \frac{3rw}{rw}\\\\V_A = \frac{V}{3}[/tex]

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