Answer:

**Answer:**

(a) 1.21 m/s

(b) 2303.33 J, 152.27 J

**Explanation:**

m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s

(a) Let their velocity after striking is v.

By use of conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v

v = ( - 356.25 + 607.94) / 208 = 1.21 m /s

(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2

= 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)

= 0.5 (1335.94 + 3270.7) = 2303.33 J

Kinetic energy after collision = 1/2 (m1 + m2) v^2

= 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J

A 0.120 kg baseball, thrown with a speed of 38.4 m/s, is hit straight back at the pitcher with a speed of 47.3 m/s. (a) What is the magnitude of the impulse delivered by the bat to the baseball? kg · m/s (b) Find the magnitude of the average force exerted by the bat on the ball if the two are in contact for 2.20 10-3 s. kN

Which sling can the crane use to lift the 1000kg pipe?A.800kg rated slingB. 1000kg rated slingC. 2000kg rated slingD. Band C

What problem did Katherine face in checking other people’s calculations in the movie ¨Hidden figures¨

Dawn is trying to find out how much weight she can push across the room. She is really trying to find her

A boy flies a kite with the string at a 30∘ angle to the horizontal. The tension in the string is 4.5 N. Part A Part complete How much work does the string do on the boy if the boy stands still?

Which sling can the crane use to lift the 1000kg pipe?A.800kg rated slingB. 1000kg rated slingC. 2000kg rated slingD. Band C

What problem did Katherine face in checking other people’s calculations in the movie ¨Hidden figures¨

Dawn is trying to find out how much weight she can push across the room. She is really trying to find her

A boy flies a kite with the string at a 30∘ angle to the horizontal. The tension in the string is 4.5 N. Part A Part complete How much work does the string do on the boy if the boy stands still?

I need a good phisics teacher for one day online please123

The **tension force **that the **rope **exerts on **block B **is **62.3 N**, the **tension force **that the rope exerts on **block A **is **1.89 N**, and the **moment **of **inertia **of the **pulley **for **rotation **about the **axle **on which it is mounted is .

Given :

- Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the free end of the rope.
- The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the pulley rotates on a frictionless axle.
- The horizontal surface on which block A (mass 2.10 kg) moves is frictionless.
- The system is released from rest, and block B (mass 7.00 kg) moves downward 1.80 m in 2.00 s.

a) First, determine the **acceleration **of the **B block**.

Now, apply Newton's second law of motion in order to determine the **tension **force that the rope exerts on **block B**.

b) Now, again apply Newton's second law of motion in order to determine the **tension **force that the rope exerts on **block A**.

c) The **sum **of the **torque **in order to determine the **moment **of **inertia **of the **pulley **for **rotation **about the axle on which it is mounted.

Now, **substitute **the **values **of the **known terms **in the above expression.

For more information, refer to the link given below:

**Answer:**

(a) 62.3 N

(b) 1.89 N

(c) 0.430 kg m²

**Explanation:**

(a) Find the acceleration of block B.

Δy = v₀ t + ½ at²

1.80 m = (0 m/s) (2.00 s) + ½ a (2.00 s)²

a = 0.90 m/s²

Draw a free body diagram of block B. There are two forces:

Weight force mg pulling down,

and tension force Tb pulling up.

Sum of forces in the -y direction:

∑F = ma

mg − Tb = ma

Tb = m (g − a)

Tb = (7.00 kg) (9.8 m/s² − 0.90 m/s²)

Tb = 62.3 N

(b) Draw a free body diagram of block A. There are three forces:

Weight force mg pulling down,

Normal force N pushing up,

and tension force Ta pulling right.

Sum of forces in the +x direction:

∑F = ma

Ta = ma

Ta = (2.10 kg) (0.90 m/s²)

Ta = 1.89 N

(c) Draw a free body diagram of the pulley. There are two forces:

Tension force Tb pulling down,

and tension force Ta pulling left.

Sum of torques in the clockwise direction:

∑τ = Iα

Tb r − Ta r = Iα

(Tb − Ta) r = I (a/r)

I = (Tb − Ta) r² / a

I = (62.3 N − 1.89 N) (0.080 m)² / (0.90 m/s²)

I = 0.430 kg m²

True

False

False. It does repeat itself

b. In the reference frame of a cosmic ray how wide does Earth seem perpendicular to the flight direction?

Express your answer with the appropriate units.

**Answer:**

6052114.67492 m

**Explanation:**

v = Velocity of cosmic ray = 0.88c

c = Speed of light =

d = Width of Earth = Diameter of Earth =

When the cosmic ray is moving towards Earth then in the frame of the cosmic ray the width of the Earth appears smaller than the original

This happens due to length contraction

Length contraction is given by

**The Earth's width is 6052114.67492 m**

Contraction only occurs in the cosmic ray's frame of reference in the direction of the ray. But in perpendicular direction the width remains unchanged.

**Hence, the width is**

A. A football flying through the air

B. An apple falling from a tree

C. A pencil rolling on the ground

D.A rocket dropping from its maximum height

A. football flying through the air

A cause it flying through the also a projectile is a object flying in the air like a arrow for example/also can I get Brainly

**Answer:**

Volume of gasoline spills out is 0.943 L.

**Explanation:**

Volumetric expansion of both gasoline and steel tank is :

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We know expansion due to temperature change is :

For gasoline:

Similarly for Steel tank:

.

Now, volume of gasoline spills out is equal to difference between expansion in volume.