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3 years ago

a ball of 150 grams is hit with a bat the ball starts travelling with a velocity of 3m/s what is the momentum of the ball

Physics

1 answer:

dangina [55]3 years ago

5 0

Answer:

150 g= 0.15 kg

Momentum = mass * velocity

= 0.15 * 3

= 0.45 kg m/s

Explanation:

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You are given three pieces of wire that have different shapes (dimensions). You connect each piece of wire separately to a batte

VikaD [51]

Answer:

$R_3 < R_1 < R_2$

Explanation:

The resistance of a wire is given by:

$R=\frac{\rho L}{A}$

where

$\rho$ is the resistivity of the material

L is the length of the wire

A is the cross-sectional area of the wire

1) The first wire has length L and cross-sectional area A. So, its resistance is:

$R_1=\frac{\rho L}{A}$

2) The second wire has length twice the first one: 2L, and same thickness, A. So its resistance is

$R_2=\frac{2\rho L}{A}$

3) The third wire has length L (as the first one), but twice cross sectional area, 2A. So, its resistance is

$R_3=\frac{\rho L}{2A}$

By comparing the three expressions, we find

$R_3 < R_1 < R_2$

So, this is the ranking of the wire from most current (least resistance) to least current (most resistance).

5 0

3 years ago

tester [92]

The pressure exerted on the block on the ground in N/m² is 200N/m².

<h3>What is pressure?</h3>

The pressure is the amount of force applied per unit area.

Given is a 5000 Newton block rests on the ground over 25 m² of area.

Pressure p = Force/Area

Put the values, we get

p = 5000 /25

p = 200 N/m²

Hence, pressure exerted on the block on the ground is 200 N/m².

Learn more about pressure.

brainly.com/question/12971272

#SPJ1

4 0

2 years ago

A toddler pushes his 7.0 kg toy box at a relatively constant velocity across the tiled floor of the family room applying a horiz

xxMikexx [17]

Answer:The coefficient of friction between the box and the floor, = 1.456 × 10⁻²

Explanation:

8 0

3 years ago

A puck of mass 0.70 kg approaches a second, identical puck that is stationary on frictionless ice. The initial speed of the movi

natali 33 [55]

Answer:

$v_1 = \ 5.196 \frac{m}{s}$

$v_2 = 3 \frac{m}{s}$

Explanation:

For this problem, we just need to remember conservation of momentum, as there are no external forces in the horizontal direction:

$\vec{p}_i = \vec{p}_f$

where the suffix i means initial, and the suffix f means final.

The initial momentum will be:

$\vec{p}_i = m_1 \ \vec{v}_{1_i} + m_2 \ \vec{v}_{2_i}$

as the second puck is initially at rest:

$\vec{v}_{2_i} = 0$

Using the unit vector $\vec{i}$ pointing in the original line of motion:

$\vec{v}_{1_i} = 6.0 \frac{m}{s} \hat{i}$

$\vec{p}_i = 0.70 \ kg \ 6.0 \frac{m}{s} \ \hat{i} + 0.70 \ kg \ 0$

$\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i}$

So:

$\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i} = \vec{p}_f$

$\vec{p}_f = 4.2 \ \frac{kg \ m}{s} \ \hat{i}$

Knowing the magnitude and directions relative to the x axis, we can find Cartesian representation of the vectors using the formula

$\ \vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

So, our velocity vectors will be:

$\vec{v}_{1_f} = v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ )$

$\vec{v}_{2_f} = v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )$

We got

$\vec{p}_f = 0.7 \ kg \ \vec{v}_{1_f} + 0.7 \ kg \ \vec{v}_{2_f}$

$4.2 \ \frac{kg \ m}{s} \ \hat{i} = 0.7 \ kg \ v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ ) + 0.7 \ kg \ v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )$

So, we got the equations:

$4.2 \ \frac{kg \ m}{s} = 0.7 \ kg \ v_1 \ cos(30 \°) + 0.7 \ kg \ v_2 \ cos(-60 \°)$

and

$0 = 0.7 \ kg \ v_1 \ sin(30 \°) + 0.7 \ kg \ v_2 \ sin(-60 \°)$ .

From the last one, we get:

$0 = 0.7 \ kg \ ( v_1 \ sin(30 \°) + \ v_2 \ sin(-60 \°) )$

$0 = v_1 \ sin(30 \°) + \ v_2 \ sin(-60 \°)$

$v_1 \ sin(30 \°) = - \ v_2 \ sin(-60 \°)$

$v_1 = \ v_2 \ \frac{sin(60 \°)}{ sin(30 \°) }$

and, for the first one:

$4.2 \ \frac{kg \ m}{s} = 0.7 \ kg \ ( v_1 \ cos(30 \°) + v_2 \ cos(60 \°) )$

$\frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} = v_1 \ cos(30 \°) + v_2 \ cos(60 \°)$

$6 \ \frac{m}{s} = (\ v_2 \ \frac{sin(60 \°)}{ sin(30 \°) } ) \ cos(30 \°) + v_2 \ cos(60 \°)$

$6 \ \frac{m}{s} = v_2 (\ \frac{sin(60 \°)}{ sin(30 \°) } ) \ cos(30 \°) + cos(60 \°)$

$6 \ \frac{m}{s} = v_2 * 2$

so:

$v_2 = 6 \ \frac{m}{s} / 2 = 3 \frac{m}{s}$

and

$v_1 = \ 3 \frac{m}{s} \ \frac{sin(60 \°)}{ sin(30 \°) }$

$v_1 = \ 5.196 \frac{m}{s}$

3 0

4 years ago

three letters (JET) are placed in front of a plane mirror the image formed is in what arrangement???

mario62 [17]

Answer:

TEJ as this is a thing you wont get

3 0

3 years ago