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Shkiper50 [21]
2 years ago
13

A 3kg book falls from a shelf. If it lands with a speed of 4.8 m/s, from what height did it fall? A. 28.8 m , B. 1.2 m , C. 0.6

m , D. 14.1 m
Physics
1 answer:
Irina-Kira [14]2 years ago
7 0
The correct answer is B. 1.2 m (apex)
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Two friends of different masses are on the playground. They are playing on the seesaw and are able to balance it even though the
Westkost [7]

Answer:

They are able to balance torques due to gravity.

F_1 L_1 = F_2L_2

Explanation:

When two friends of different masses will balance themselves on see saw then at equilibrium position the see saw will remain horizontal

This condition will be torque equilibrium position where the see saw will not rotate

Here we can say

F_1 L_1 = F_2L_2

here we know that force is due to weight of two friends

and their positions are different with respect to the lever about which see saw is rotating

since both friends are of different weight so they will balance themselves are different positions as per above equation

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How many electrons are in the third energy level of a calcium atom
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A pitcher throws a baseball with a mass of 143 g horizontally at a speed of 38.8 m/s (87 mi/h). The hitter's bat is in contact w
SIZIF [17.4K]

Answer:

F = −10093.41 N

Explanation:

Given that,

Mass of a baseball, m = 143 g = 0.143 kg

Initial speed of the baseball, u = +38.8 m/s

The hitter's bat is in contact with the ball for 1.20 ms and then travels straight back to the pitcher's mound at a speed of 45.9 m/s, v = -45.9 m/s

We need to find the average force exerted on the ball by the bat. So, Force is given by :

F=ma

a is acceleration

F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.143\times (-45.9-(38.8))}{1.2\times 10^{-3}}\\\\F=-10093.41\ N

So, the average force exerted on the ball by the bat has a magnitude of 10093.41 N.

8 0
3 years ago
a tiger moving with constant accleration covers the distance between two points 70 meter apart in 7 seconds . its speed as it pa
makvit [3.9K]

Answer:

a = 1.428 [m/s²]

v₀ = 5 [m/s]

Explanation:

To solve this problem we must use the following equation of kinematics.

x=x_{o}+v_{o}*t+\frac{1}{2}*a*t^{2}

where:

x = final point [m]

x₀ = initial point [m]

v₀ = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

But we need to use this additional equation.

v_{f}=v_{o}+a*t

where:

vf = final velocity = 15 [m/s]

Now we can use this equation, replacing it, in the first one. We must bear in mind that the difference among x - x₀ is equal to 70 [m]

x-x_{o}=v_{o}*t+\frac{1}{2}*a*t^{2} \\x-x_{o}=(v_{f}-a*t)*t+\frac{1}{2} *a*t^{2}\\70=(15-a*t)*t+\frac{1}{2}*a*t^{2}\\70=15*t-a*t^{2} +\frac{1}{2}*a*t^{2} \\70=15*t-\frac{1}{2}*a*t^{2}\\70=15*(7)-\frac{1}{2} *a*(7)^{2}\\105-70=0.5*a*49\\35=24.5*a\\a=1.428[m/s^{2} ]

Now replacing this value in the second equation, we can find the initial velocity.

15=v_{o}+1.428*7\\v_{o}=5[m/s]

3 0
3 years ago
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