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Sladkaya [172]
2 years ago
8

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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A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.60m and the horizontal
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Answer:

v = 46.99 m/s

Explanation:

The velocity of the ball just before it touches the ground, is given by the following formula:

v=\sqrt{v_x^2+v_y^2}           (1)

vx: horizontal component of the velocity

vy: vertical component of the velocity

The vertical component vy is calculated by using the following formula:

v_y^2=v_{oy}^2+2gh   (2)

vy: final velocity

voy: initial vertilal velocity = 0m/s  (because it is a semi parabolic motion)

g: gravitational acceleration = 9.8 m/s^2

h: height = 1.60m

You replace the values of the parameters in the equation (2):

v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}

vx is calculated by using the information about the horizontal range of the ball:

R=v_o\sqrt{\frac{2h}{g}}    (3)

R: horizontal range of the ball = 20.0 m

You solve the previous equation for vo, the initial horizontal velocity:

v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}

The horizontal component of the velocity is constant in the complete trajectory, hence, you have that

vx = vo = 35 m/s

Finally, you replace the values of vx and vy in the equation (1):

v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}

The velocity of the ball just before it touches the ground is 46.99 m/s

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Protons and neutrons are packed together in a very small region called nucleus. Protons are positively charged and we know that like charges repel. Then how is it that protons are not repelling each other and flying away from nucleus?

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