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Ivan
3 years ago
7

A box of mass 1 kg is hung from a spring scale. The reading on the spring scale is approximately 10N. What would be the reading

on the spring scale if a box of mass 5 kg is hung from it? (Assume acceleration due to gravity to be 10 m/s2.)
A) 5 N
B) 10 N
C) 50 N
D) 500 N
Physics
2 answers:
8090 [49]3 years ago
8 0
The answer will be 50N.
This is because the spring reads weight and weight is mass times acceleration due to gravity.5kg*10m/s2=50N
Blababa [14]3 years ago
6 0

Explanation:

It is given that,

When mass is, m₁ = 1 kg

The reading on the spring scale, F = 10 N

If the mass is, m₂ = 5 kg

We need to find the reading on the spring scale. The force of gravity is acting on the box which is given by :  

F=mg

F=5\ kg\times 10\ m/s^2

F = 50 N

So, the reading on the spring scale is 50 N. Hence, this is the required solution.                            

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A baseball is hit that just goes over a wall that is 45.4m high. If the baseball is traveling at 46.2 m/s at an angle of 32.7° b
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Answer:

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If the ball has a total velocity of 46.2 m/s, at an angle of -32.7 degrees, we can decompose its speed into its horizontal and vertical components.

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Since a ball thrown is at free fall, only affected by gravity (omitting air drag), we can say it is affected by constant acceleration, therefore we can use

Y(t) = Y0 + Vy0 *t + 1/2 * a * t^2

We consider t=0 as the moment when the ball was hit, so in this case Y0 = 1 m

If we take the first derivative of the equation of position, we get the equation for speed

V(t) = Vy0 + a * t

We know that being t2 the moment the ball goes over the wall

V(t2) = -25 m/s

Y(t2) = 45.4 m

So:

45.4 = 1 + Vy0 * t2 + 1/2 * a * t2^2

-25 = Vy0 + a * t2

Then:

Vy0 = -25 - a * t2

So:

45.4 = 1 + (-25 - a * t2) * t2 + 1/2 * a * t2^2

0 = -44.4 - 25 * t2 - 1/2 * a * t2^2

a = -9.81 m/s^2

0 = -44.4 - 25 * t2 + 4.9 * t2^2

Solving this quadratic equation we get:

t1 = -1.39 s

t2 = 6.5 s

Since we are looking for a positive value we disregard t1.

Now we can obtain Vy0:

Vy0 = -25 + 9.81 * 6.5 = 38.76 m/s

Since horizontal speed is constant Vx0 = 38.9 m/s

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a = atan(Vy0/Vx0) = 44.9 degrees

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