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2 years ago

A train moves at 300 km/h. How far will it go in 5 hours?

Physics

1 answer:

faust18 [17]2 years ago

5 0

Answer:

1,500 km

Explanation:

300 km/h

300 kilometers PER HOUR

So you want to see how far it will go in 5 hours, all you need to do is multiply the kilometers by 5. Since we know that per hour it's 300 km.

300 * 5 = 1,500

The train moves at 1,500 km in 5 hours.

Hope this helped!

Have a supercalifragilisticexpialidocious day!

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A homing pigeon starts from rest and accelerates uniformly at +4.00 m/s squared for 10.0 seconds. What is its velocity after the

zavuch27 [327]

' +4 m/s² ' means that the pigeon's speed is 4 m/s greater every second.

Starting from zero speed, after 10 seconds, its speed is

(10 x 4m/s) = 40 m/s.

We can't say anything about its velocity, because we have
no information regarding the direction of its flight.

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3 years ago

A tennis player smashes a ball of mass m horizontally at a vertical wall. The ball rebounds at the same speed v with which it st

Morgarella [4.7K]

Answer:

B.0

Explanation:

Change in momentum: This is defined as the product of mass and change in velocity of a body. or it can be defined as the product of force and time of a body. The fundamental unit of change in momentum is kg.m/s

Change in momentum = M(V-U)......................... Equation 1

where M = mass of the ball, V = final velocity of the ball, U = initial velocity of the ball.

Let: M = m kg and V = U = v m/s

Substituting these values into equation 1

Change in momentum = m(v-v)

Change in momentum = m(0)

Change in momentum = 0 kg.m/s

Therefore the momentum of the ball has not changed.

The right option is B.0

5 0

2 years ago

A system gains 757 kJ757 kJ of heat, resulting in a change in internal energy of the system equal to +176 kJ.+176 kJ. How much w

Alecsey [184]

Answer:

581 kJ, work was done by the system

Explanation:

According to the first law of thermodynamics:

$\Delta U = Q-W$

where

$\Delta U$ is the change in internal energy of the system

Q is the heat absorbed by the system (positive if absorbed, negative if released)

W is the work done by the system (positive if done by the system, negative if done by the surrounding)

In this problem,

$Q=+757 kJ$

$\Delta U = +176 kJ$

Therefore the work done by the system is

$W=Q-\Delta U=757 kJ-176 kJ=+581 kJ$

And the positive sign means the work is done BY the system.

5 0

2 years ago

A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur

muminat

Answer:

a) E = 0

b) $E = \dfrac{k_e \cdot q}{ r^2 }$

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

From which we have;

$E \cdot A = \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}} = \dfrac{0}{\varepsilon _{0}} = 0$

E = 0/A = 0

E = 0

b) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

$E \cdot A = \dfrac{+q }{\varepsilon _{0}}$

$E = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}$

By Gauss theorem, we have;

$E\oint dS = \dfrac{q}{\varepsilon _{0}}$

Therefore, we get;

$E \cdot (4 \cdot \pi \cdot r^2) = \dfrac{q}{\varepsilon _{0}}$

The electrical field outside the spherical shell

$E = \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }= \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }$

$k_e= \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }$

Therefore, we have;

$E = \dfrac{k_e \cdot q}{ r^2 }$

5 0

2 years ago

•• CP Two blocks connected by a light horizontal rope sit at rest on a horizontal, frictionless surface. Block AA has mass 15.0

Firdavs [7]

Answer:

(a) T= 38.4 N

(b) m= 26.67 kg

Explanation:

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Kinematics

d= v₀t+ (1/2)*a*t² (Formula 2)

d:displacement in meters (m)

t : time in seconds (s)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

v₀=0, d=18 m , t=5 s

We apply the formula 2 to calculate the accelerations of the blocks:

d= v₀t+ (1/2)*a*t²

18= 0+ (1/2)*a*(5)²

a= (2*18) / ( 25) = 1.44 m/s² to the right

We apply Newton's second law to the block A

∑Fx = m*ax

60-T = 15*1.44

60 - 15*1.44 = T

T = 38.4 N

We apply Newton's second law to the block B

∑Fx = m*ax

T = m*ax

38.4 = m*1.44

m= (38.4) / (1.44)

m = 26.67 kg

7 0

3 years ago