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2 years ago

A train travels at 270 km in 3 hours . what is the speed of the train?

Physics

2 answers:

yKpoI14uk [10]2 years ago

7 0

Rate times time = distance.

Rate = x
Time = 3 hours
Distance =270 km

Ann [662]2 years ago

3 0

$270 \div 3 = 90km / h$

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In a nuclear station if the temperature of the superheated water starts to decrease what should the engineer in charge of the re

Sophie [7]

Answer:

They sh0uld g0 t0 the reactor and then, see what the issue is...

Explanation:

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2 years ago

A projectile is fired at an upward angle of 45.0º from the top of a 265-m cliff with a speed of .sm 185 What will be its speed w

nignag [31]

The final velocity of the projectile when it strikes the ground below is 198.51 m/s.

<h3>Time of motion of the projectile</h3>

The time taken for the projectile to fall to the ground is calculated as follows;

h = vt + ¹/₂gt²

where;

h is height of the cliff
v is velocity
t is time of motion

265 = (185 x sin45)t + (0.5)(9.8)t²

265 = 130.8t + 4.9t²

4.9t² + 130.8t - 265 = 0

solve the quadratic equation using formula method,

t = 1.89 s

<h3>Final velocity of the projectile</h3>

vyf = vyi + gt

where;

vyf is the final vertical velocity
vyi is initial vertical velocity

vyf = (185 x sin45) + (9.8 x 1.89)

vyf = 149.322 m/s

vxf = vxi

where;

vxf is the final horizontal velocity
vxi is the initial horizontal velocity

vxf = 185 x cos(45)

vxf = 130.8 m/s

vf = √(vyf² + vxf²)

where;

vf is the speed of the projectile when it strikes the ground below

vf = √(149.322² + 130.8²)

vf = 198.51 m/s

Learn more about final velocity here: brainly.com/question/6504879

#SPJ1

3 0

1 year ago

There are two uncharged objects, A and B. Some electrons move from A to B. How does the charge of the two objects change?

miskamm [114]

A becomes positive, while b is now negative. Basically, electrons are negative particles. If they go to somewhere, they make the somewhere they go to negative.

5 0

3 years ago

How much kinetic energy does a system containing 3 moles of an ideal gas at 300 K possess? What is the heat capacity at constant

ElenaW [278]

Answer:

1. K.E = 11.2239 kJ ≈ 11.224 kJ

2. $C_{V} = 37.413 JK^{- 1}$

3. $Q = 10.7749 kJ$

Solution:

Now, the kinetic energy of an ideal gas per mole is given by:

K.E = $\frac{3}{2}mRT$

where

m = no. of moles = 3

R = Rydberg's constant = 8.314 J/mol.K

Temperature, T = 300 K

Therefore,

K.E = $\frac{3}{2}\times 3\times 8.314\times 300$

K.E = 11223.9 J = 11.2239 kJ ≈ 11.224 kJ

Now,

The heat capacity at constant volume is:

$C_{V} = \frac{3}{2}mR$

$C_{V} = \frac{3}{2}\times 3\times 8.314 = 37.413 JK^{- 1}$

Now,

Required heat transfer to raise the temperature by $15^{\circ}$ is:

$Q = C_{V}\Delta T$

$\Delta T = 15^{\circ} = 273 + 15 = 288 K$

$Q = 37.413\times 288 = 10774.9 J = 10.7749 kJ$

8 0

2 years ago

Two carts, one twice as heavy as the other, are at rest on a horizontal track. A person pushes each cart for 8 s. Ignoring frict

GalinKa [24]

Answer:

The correct option is: B that is 1/2 K

Explanation:

Given:

Two carts of different masses, same force were applied for same duration of time.

Mass of the lighter cart = $m$

Mass of the heavier cart = $2m$

We have to find the relationship between their kinetic energy:

Let the KE of cart having mass m be "K".

and KE of cart having mass m be "K1".

As it is given regarding Force and time so we have to bring in picture the concept of momentum Δp and find a relation with KE.

Numerical analysis.

⇒ $KE = \frac{mv^2}{2}$

⇒ $KE = \frac{mv^2}{2}\times \frac{m}{m}$

⇒ $KE = \frac{m^2v^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(mv)^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(\triangle p)^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(\triangle p)^2}{2m}=\frac{(F\times t)^2}{2m}$

Now,

Kinetic energies and their ratios in terms of momentum or impulse.

KE (K) of mass m.

⇒ $K=\frac{(F\times t)^2}{2m}$ ...equation (i)

KE (K1) of mass 2m.

⇒ $K_1=\frac{(F\times t)^2}{2\times 2m}$

⇒ $K_1=\frac{(F\times t)^2}{4m}$ ...equation (ii)

Lets divide K1 and K to find the relationship between the two carts's KE.

⇒ $\frac{K_1}{K} =\frac{(F\times t)^2}{4m} \times \frac{2m}{(F\times t)^2}$

⇒ $\frac{K_1}{K} =\frac{2m}{4m}$

⇒ $\frac{K_1}{K} =\frac{2}{4}$

⇒ $\frac{K_1}{K} =\frac{1}{2}$

⇒ $K_1=\frac{K}{2}$

⇒ $K_1=\frac{1}{2}K$

The kinetic energy of the heavy cart after the push compared to the kinetic energy of the light cart is 1/2 K.

7 0

2 years ago