A person rolls a 7 kg bowling ball down a lane in a bowling alley. The lane is
18 m long. The ball is traveling at 7 m/s when it leaves the person's hand.
What is the ball's kinetic energy at this point?

Answer:

The speed of the combined mass after the collision is 2.1 m/s.

Explanation:

mass of runner, m = 70 kg

speed  of runner, u = 2.7 m/s

mass of shortstop, m' = 85 kg

speed  of shortstop, u' = 1.6 m/s

Let the velocity of combined system is v.

Use conservation of momentum

Momentum before collision = momentum after collision

m u + m' u' = (m + m') v

70 x 2.7 + 85 x 1.6 = (70 + 85) v

189 + 136 = 155 v

v = 2.1 m/s

Answer:

B

Explanation:

The seasons are measured in how far or close the earth is to the sun.

Answer:

The efficiency of Carnot's heat engine is 26.8 %.

Explanation:

Temperature of hot reservoir, TH = 100 degree C = 373 K

temperature of cold reservoir, Tc = 0 degree C = 273 K

The efficiency of Carnot's heat engine is

[tex]\eta = 1-\frac{Tc}{T_H}\\\\\eta = 1 -\frac{273}{373}\\\\\eta = 0.268 =26.8 %[/tex]

The efficiency of Carnot's heat engine is 26.8 %.

This question is incomplete, the missing diagram is uploaded along this answer below.

Answer:

- the magnitude of the tension force exerted by the hamstring muscles on the leg is 990 N

- the magnitude of compression force at the knee joint is 900 N

Explanation:

Given the data in the question and diagram below;

Net torque = 0

Torque = force × lever arm

so

F[tex]_{ConF[/tex]  × ( 15.0 in + 1.5 in ) = T[tex]_{HonL[/tex] × 1.5 in

given that F[tex]_{ConF[/tex] = 90 N

90 × ( 15.0 in + 1.5 in ) = T[tex]_{HonL[/tex] × 1.5 in

90 N × 16.5 in =  T[tex]_{HonL[/tex] × 1.5 in

T[tex]_{HonL[/tex] = ( 90 N × 16.5 in ) / 1.5 in

T[tex]_{HonL[/tex] = 990 N

Therefore, the magnitude of the tension force exerted by the hamstring muscles on the leg is 990 N

b) magnitude of compression force at the knee joint;

In equilibrium, net force = 0

along horizontal

F[tex]_{FonB[/tex] - T[tex]_{HonL[/tex] + F[tex]_{ConF[/tex] = 0

we substitute

F[tex]_{FonB[/tex] - 990 + 90 = 0

F[tex]_{FonB[/tex] - 900 = 0

F[tex]_{FonB[/tex] = 900 N

Therefore, the magnitude of compression force at the knee joint is 900 N

Answer:

Explanation:

Here's what we know and in which dimension:

y dimension:

[tex]v_0=30[/tex] m/s

v = 0 (I'll get to that injust a second)

a = -9.8 m/s/s

The final velocity of 0 is important because that's the velocity of the ball right at the very top of its travels. If we knew how long it takes to get to that max height, we can also use that to find out how long it will take to hit the ground. Therefore, we will find the time it takes to reach its max height and pick up with the investigation of what this means after.

x dimension:

Δx = 70 m

v = ??

Velocity is our unknown.

Solving for the time in the y dimension:

[tex]v=v_0+at[/tex] and filling in:

0 = 30 + (-9.8)t and

-30 = -9.8t so

t = 3.1 seconds

We know it takes 3.1 seconds to get to its max height. In order to determine how long it will take to hit the ground, just double the time. Therefore, it will take 6.2 seconds for the ball to come back to the ground, which is where the persom trying to catch the ball comes in. We will use that time in our x dimension now.

In the x dimension, the equation we need is just a glorified d = rt equation since the acceleration in this dimension is 0.

Δx = vt and

70 = v(6.2) so

v = 11.3 m/s

Answer:

M L1 = m L2       torques must be zero around the fulcrum

M = m L2 / L1 = .3 kg * 28 cm / 22 cm = .382 kg

Answer:

The right solution is "50200 days".

Explanation:

Given:

Calories intake,

= 6000 kcal,

or,

= [tex]2.52\times 10^7 \ J[/tex]

Force,

= 500 N

As we know,

⇒ [tex]Work \ done = Force\times distance[/tex]

Or,

⇒ [tex]distance = \frac{Work \ done}{Force}[/tex]

By putting the values, we get

                  [tex]=\frac{2.52\times 10^7}{500}[/tex]

                  [tex]=0.502\times 10^5[/tex]

                  [tex]=50200 \ m[/tex]

hence,

The number of days will be:

= [tex]\frac{50200}{1}[/tex]

= [tex]50200 \ days[/tex]

Answer:

1) 0.3675

2) 0.367

3) 72.6186

4) 0.5

5) 0.000671

Answer:

1) 367.5 mg = 0.3675 g

2) 367 mL = 0.367 L

3) 28.59 in = 72.61 cm

4) 8 0z = 0.5 lb

5) 0.671 mm = 0.0000671 m

Explanation:

Given that,

35 m/s at 57° from the x-axis.

Speed, v = 35 m/s

Angle, θ = 57°

Horizontal component,

[tex]v_x=v\cos\theta\\\\=35\times \cos(57)\\\\=19.06 m/s[/tex]

Vertical component,

[tex]v_y=v\sin\theta\\\\v_y=35\times \sin(57)\\\\=29.35\ m/s[/tex]

Hence, this is the required solution.

Answer:

12+2=24+30+2=66

Explanation:

Answer:

The correct answer is "4000 J".

Explanation:

Given that,

Force,

= 200 N

Displacement,

= 20 m

Now,

The work done will be:

⇒ [tex]Work=Force\times displacement[/tex]

By putting the values, we get

              [tex]=200\times 20[/tex]

              [tex]=4000 \ J[/tex]

Complete question is;

A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h and g:

(a) the work done by the force of gravity on the projectile,

(b) the change in kinetic energy of the projectile since it was fired, and

(c) the final kinetic energy of the projectile.

(d) Are any of the answers changed if the initial angle is changed?

Answer:

A) W = mgh

B) ΔKE = mgh

C) K2 = mgh + ½mv_o²

D) No they wouldn't change

Explanation:

We are expressing in terms of m, v0​, h, and g. They are;

m is mass

v0 is initial velocity

h is height of projectile fired

g is acceleration due to gravity

A) Now, the formula for workdone by force of gravity on projectile is;

W = F × h

Now, Force(F) can be expressed as mg since it is force of gravity.

Thus; W = mgh

Now, there is no mention of any angles of being fired because we are just told it was fired horizontally.

Therefore, even if the angle is changed, workdone will not change because the equation doesn't depend on the angle.

B) Change in kinetic energy is simply;

ΔKE = K2 - K1

Where K2 is final kinetic energy and K1 is initial kinetic energy.

However, from conservation of energy, we now that change in kinetic energy = change in potential energy.

Thus;

ΔKE = ΔPE

ΔPE = U2 - U1

U2 is final potential energy = mgh

U1 is initial potential energy = mg(0) = 0. 0 was used as h because at initial point no height had been covered.

Thus;

ΔKE = ΔPE = mgh

Again like a above, the change in kinetic energy will not change because the equation doesn't depend on the angle.

C) As seen in B above,

ΔKE = ΔPE

Thus;

½mv² - ½mv_o² = mgh

Where final kinetic energy, K2 = ½mv²

And initial kinetic energy = ½mv_o²

Thus;

K2 = mgh + ½mv_o²

Similar to a and B above, this will not change even if initial angle is changed

D) All of the answers wouldn't change because their equations don't depend on the angle.

Answer:

So the answer is B. A is wrong because negative answer = deceleration

Answer:

the horizontal component of the force is 50 N

Explanation:

Given;

force applied by the man, F = 100 N

angle of inclination of the force, θ = 60⁰

mass of the dog, m = 20 kg

The horizontal component of the force is calculated as;

[tex]F_x = F\times cos(\theta)\\\\F_x = 100 \ N \times cos(60^0)\\\\F_x = 100\ N \times 0.5\\\\F_x = 50 \ N[/tex]

Therefore, the horizontal component of the force is 50 N

We call the acceleration of an object moving in uniform circular motion— resulting from a net external force—the centripetal ...

Answer:

the electric field is  3.91 x 10⁶ N/C

Explanation:

Given the data in the question;

Electric field at a point due to point charge is;

E = kq/r²

where k is the constant, r is the distance from centre of terminal to point where electric field is, q is the excess charge placed on the centre of terminal of Van de Graff,a generator

Now, given that r = 3.9 m, k = 9.0 x 10⁹ Nm²/C², q = 6.60 mC = 6.60 x 10⁻³ C

so we substitute into the formula

E = [(9.0 x 10⁹ Nm²/C²)( 6.60 x 10⁻³ C)] / ( 3.9 )²

E = 59400000 / 15.21

E = 3.91 x 10⁶ N/C

Therefore, the electric field is  3.91 x 10⁶ N/C

Answer: The weight of the object is 29.4 N

Explanation:

To calculate the weight of the object, we use the equation:

[tex]W=m\times g[/tex]

where,

m = mass of the object = 3 kg

g = acceleration due to gravity = [tex]9.8m/s^2[/tex]

Putting values in above equation, we get:

[tex]W=3kg\times 9.8m/s^2\\\\W=29.4N[/tex]

Hence, the weight of the object is 29.4 N

Answer:

Refer to the attachment!~

Explanation:

Energy input = F×d = (150 N)(5.5 m) = 825 J

Energy output = mgh = (66 kg)(9.8 m/s^2)(1.10 m) = 711 J

efficiency = [tex]\dfrac{\text{output}}{\text{input}}[/tex]×100% = 86.2%

Answer:

30

Explanation:

since  [tex]\frac{30}{5}[/tex]=6

         [tex]\frac{30}{6}[/tex]=5

then both 5 and 6 are factors of 30

Have a nice day

Answer:

559.5 N

Explanation:

Applying,

v² = u²+2gs............. Equation 1

Where v = final velocity,

From the question,

Given: s = 5.10 m, u = 0 m/s ( from rest)

Constant: 9.8 m/s²

Therefore,

v² = 0²+2×9.8×5.1

v² = 99.96

v = √(99.96)

v = 9.99 m/s

As the diver eneters the water,

u = 9.99 m/s, v = 0 m/s

Given: t = 1.34 s

Apply

a = (v-u)/t

a = 9.99/1.34

a = -7.46 m/s²

F = ma.............. Equation 2

Where F = force, m = mass

Given: m = 75 kg, a = -7.46 m/s²,

F = 75(-7.46)

F = -559.5 N

Hence the average force exerted on the diver is 559.5 N

Answer:

less than the weight of the block.

Explanation:

From the free body diagram, we get.

The normal force is N = Mg cosθ

The tension in the string is T = Mg sinθ

Wight of the block when the block is static, W = Mg

Now since the magnitude of cosθ is in the range of : 0 < cosθ < 1,

therefore, the normal force is less than the weight of the static block.

Answer:

A

Explanation:

I guess not that much confidential!

Answer:

Mm, thats the answer trust me men

Answer:

By measure the effective length and the time period of the pendulum.

Explanation:

Let the student take the oscillating pendulum at the planet.

He measure the time period of the pendulum  by using the stop watch or the ordinary watch.

Then measure the effective length of the pendulum which is the distance between the center of gravity of the bob and the point of suspension of the pendulum.

Now, use the formula of the time period of the pendulum,

[tex]T =2\pi\sqrt\frac{L}{g}[/tex]

Here, L is the effective length of the pendulum, g is the acceleration due to gravity at the planet and T is time period of the pendulum.  

By rearranging the terms, we get

[tex]T =2\pi\sqrt\frac{L}{g}\\\\T^{2}=4\pi^2\times\frac{L}{g}\\\\g =\frac{4\pi^2L}{T^2}[/tex]

Here, by substituting the values of L and T, the student get the value of acceleration due to gravity at that planet.

Answer:

C. both a speed and a direction

The boy pushed the sled for 16 seconds.

We have a boy who pushes his little brother on a sled.

We have to determine for how long time does boy push the sled.

The Work - Energy theorem states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

According to the question -

The sled is initially at rest → initial velocity (u) = 0.

Final velocity (v) = 4 m/s

Mass of boy and sled (M) = 40 kg

Power developed (P) = 20 W = 20 Joules/sec

According to work - energy theorem -

Work done (W) = Δ E(K) = E(f) - E(i)

Therefore -

W = ([tex]\frac{1}{2} \times 40 \times 4 \times 4 - \frac{1}{2}[/tex] x 40 x 0) = 320 Joule

Now, Power is defined as the rate of doing work -

P = [tex]\frac{dW}{dt}[/tex] = [tex]\frac{W}{t}[/tex]

20 = [tex]\frac{320}{t}[/tex]

t = 16 seconds

Hence, the boy pushed the sled for 16 seconds.

To solve more questions on Work, Energy and Power, visit the link below -

https://brainly.com/question/208670

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