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g100num [7]
2 years ago
12

A ball rolls down an incline with an acceleration of 10 cm/s^2. If it starts with an initial velocity of 0 cm/s and has a velocy

of 50 cm/s when it reaches the bottom of the ramp, how long is the ramp?
Physics
1 answer:
11Alexandr11 [23.1K]2 years ago
8 0
Given a = 10 cm/s²
          u = 0 cm/s
          v = 50 cm/s
we know that 
         v²=u²+2aS
        2500=2×10×S
        2500÷20 = S
        S= 125 cm
The ramp is 125 cm
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Why does it takes a long time to heat a room with a high ceiling?
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Answer:

High ceilings make a room feel large and open, but they can be difficult to cool and heat. Because hot air rises, the challenge becomes trying to keep the hot air where you want it and preventing if from being wasted where you don't.

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6 0
2 years ago
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a 90 kg architect is standing 2 meters from the center of a scaffold help up by a rope on both sides. the scaffold is 6m long an
Mademuasel [1]
We can solve the problem by requiring the equilibrium of the forces and the equilibrium of torques.

1) Equilibrium of forces:
T_1 - W_p - W_s + T_2 =0
where
W_p = (90kg)(9.81 m/s^2)=883 N is the weight of the person
W_s = (200kg)(9.81 m/s^2)=1962 N is the weight of the scaffold
Re-arranging, we can write the equation as
T_1 = 2845 N-T_2 (1)

2) Equilibrium of torques:
T_1 \cdot 3 m - W_p \cdot 2 m - T_2 \cdot 3m =0
where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using W_p = 883 N and replacing T1 with (1), we find
2845 N \cdot 3 m - T_2 \cdot 3 m - 833 N \cdot 2 m - T_2 \cdot 3 m=0
from which we find
T_2 = 1128 N

And then, substituting T2 into (1), we find
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8 0
2 years ago
An electric field of 1.32 kV/m and a magnetic field of 0.516 T act on a moving electron to produce no net force. If the fields a
Lapatulllka [165]

Answer:

The speed of the electron is 2.55\times 10^3\ m/s.

Explanation:

Given that,

The magnitude of electric field, E=1.32\ kV/m=1.32\times 10^3\ V/m

The magnitude of magnetic field, B = 0.516 T

Both the magnetic and electric fields are acting on the moving electron. Then,  the magnitude of electric field and magnetic field is balanced such that :

evB=eE\\\\v=\dfrac{E}{B}\\\\v=\dfrac{1.32\times 10^3}{0.516}\\\\v=2558.13\ m/s

or

v=2.55\times 10^3\ m/s

So, the speed of the electron is 2.55\times 10^3\ m/s. Hence, this is the required solution.

3 0
3 years ago
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3 years ago
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