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3 years ago

10

Suppose you throw a ball vertically upward with a speed of 49 m/s. Neglecting air friction, what would be the height of the ball

be 2.0 second later?

1 answer:

vlabodo [156]3 years ago

6 0

Answer:

78.4 m

Explanation:

Using newton's equation of motion,

S = ut + 1/2gt²......................... Equation 1

Where S = Height, t = time, u = initial velocity, g = acceleration due to gravity.

Note: Taking upward to be negative, and down ward positive

Given: u = 49 m/s, t = 2.0 s, g = -9.8 m/s²

Substitute into equation 1

S = 49(2) - 1/2(9.8)(2)²

S = 98 - 19.6

S = 78.4 m

Hence the height of the ball two seconds later = 78.4 m

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<u><em>(b) 1:1</em></u>

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<u></u>

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A) 1 cal

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Explanation:

A)

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$Q=mC\Delta T$

where

m is the mass of the substance

C is the specific heat capacity

$\Delta T$ is the change in temperature

In this problem:

m = 1 g is the mass of water

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$\Delta T=1^{\circ}C$ is the change in temperature

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For a solid substance at its melting point, the amount of heat needed to melt completely the substance is given by

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m is the mass of the substance

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C)

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where

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So, the heat needed is

$Q=(1)(540)=540 cal$

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