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morpeh [17]
3 years ago
10

Why are experiments often performed in laboratories?

Physics
1 answer:
Marrrta [24]3 years ago
6 0

Answer:

B

Explanation:

Laboratories are usually controlled: however, in public places, there are way too many factors that could change an experiment

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A penny falls from a windowsill 25.0 m above the sidewalk. How fast is the penny moving when it strikes the ground? (Remember th
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To determine, the final velocity we use one kinematic equation. It would be 2ax = vf^2 - v0^2. The initial velocity would be zero since it starts from rest. Therefore, the equation would be:

vf = √(2ax)        where a is the acceleration ( a = g ) and x is the height

vf = √(2)(9.8)(25)
vf = - 22.14 m/s
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A car on the highway slows for 32.0 m/s to rest as the driver approaches a traffic jam. The car has 70.0. To stop. A.Solve for t
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A ball of mass 0.3 kg is released from rest at a height of 8 m. How fast is it going when it hits the ground? (Gravity being equ
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Answer:

Explanation:

Mass doesn't matter here because when something is falling, gravity plays fairly; an elephant falls at the same rate of acceleration as does a feather. What DOES matter is everything pertinent to the y-dimension of free-fall:

a = -9.8 m/s/s

v₀ = 0 (since the ball was held before it was dropped)

v = ??

Δx = -8 m (negative because the ball drops this far below the point from which it was released).

Putting all this together in one equation:

v² = v₀² + 2aΔx and filling in this equation:

v² = (0)² + 2(-9.8)(-8) and

v² = 156.8 so

v = 12.5 which rounds to 13 if you're using 2 sig figs, and rounds to 10 if you're only using 1 (which you should be, according to the way the numbers have been given in this problem)

3 0
3 years ago
A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a forc
____ [38]

Answer:

Part a)

a = 0.36 m/s^2

Part b)

a = -1.29 m/s^2

Explanation:

Force applied by the student on the box is 80 N at an angle of 25 degree

so here two components of the force on the box is given as

F_x = Fcos25

F_x = 80 cos25 = 72.5 N

F_y = Fsin25

F_y = 80 sin25 = 33.8 N

now in vertical direction we can use force balance for the box to find the normal force on it

F_n + F_y = mg

F_n = (25)(9.81) - 33.8

F_n = 211.45 N

now kinetic friction on the box opposite to applied force due to rough floor is given as

F_k = \mu F_n

F_k = (0.300)(211.45) = 63.44 N

now the net force on the box in forward direction is given as

F_{net} = F_x - F_k

F_{net} = 72.5 - 63.44 = 9.065 N

now the acceleration of the box is given as

a = \frac{F_{net}}{m}

a = \frac{9.065}{25} = 0.36 m/s^2

Part b)

when box is pulled up along the inclined surface of angle 10 degree

now the two components of the force will be same along the inclined and perpendicular to inclined plane

F_x = 72.5 N

F_y = 33.8 N

now force balance perpendicular to inclined plane is given as

F_n + F_y = mgcos\theta

F_n = (25)(9.81)cos10 - 33.8 = 207.7 N

now the friction force opposite to the motion on the box is given as

f_k = \mu F_n

f_k = (0.300)(207.7) = 62.3 N

now the net pulling force along the inclined plane is given as

F_{net} = F_x - F_k - mgsin10

F_{net} = 72.5 - 62.3 - (25)(9.81)sin10

F_{net} = -32.38 N

now the box will decelerate and it is given as

a = \frac{F_{net}}{m}

a = \frac{-32.38}{25} = -1.29 m/s^2

7 0
4 years ago
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