1. You have a cat who has a mass of 10 kg and is
1 answer:
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Answer:
40 cm
Explanation:
We are given that
Load=800 N
Effort=200 N
Load distance=10 cm
We have to find the effort distance.
We know that
Using the formula
Effort distance=
Effort distance=
Effort distance=40 cm
Hence, the effort distance will be 40 cm.
Answer:
a) Δx = 11.6 m
b) t = 3.9 s
Explanation:
a)
- Since the snowmobile is moving at constant speed, and the drive force is 195 N, this means that thereis another force equal and opposite acting on it, according to Newton's 2nd Law, due to there is no acceleration present in the horizontal direction .
- This force is just the force of kinetic friction, and is equal to -195 N (assuming the positive direction as the direction of the movement).
- Once the drive force is shut off, the only force acting on the snowmobile remains the friction force.
- According Newton's 2nd Law, this force is causing a negative acceleration (actually slowing down the snowmobile) that can be found as follows:
- Assuming the friction force keeps constant, we can use the following kinematic equation in order to find the distance traveled under this acceleration before coming to an stop, as follows:
- Taking into account that vf=0, replacing by the given (v₀) and a from (1), we can solve for Δx, as follows:
b)
- We can find the time needed to come to an stop, applying the definition of acceleration, as follows:
- Since we have already said that the snowmobile comes to an stop, this means that vf = 0.
- Replacing a and v₀ as we did in (3), we can solve for Δt as follows:
Answer:
C) 7.35*10⁶ N/C radially outward
Explanation:
- If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
- So, we can write the following equation:
- As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
- So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
- So, on the outer surface of the shell there must be a charge that be the difference between them:
- Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:
- As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.