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2 years ago

3 A picture is supported by two vertical strings; if the weighi

Physics

1 answer:

andrezito [222]2 years ago

7 0

Answer:

The force exerted by each string is 25 N

Explanation:

Net Force

The net force is the vector sum of forces acting on a body. The net force is a single force that represents the effect of the original forces on the body's motion. It gives the particle the same acceleration as all those actual forces together as described by Newton's second law of motion.

The picture described in the problem is hanging at rest supported by two vertical strings. This means that the net force acting on it is zero.

Assume the magnitude of each of these equal forces is F, and the picture has a weight of W=50 N, thus the net force is:

F + F - W

The positive signs indicate an upwards direction and the negative sign means a downwards direction. Since the net force is zero:

F + F - W = 0

2F = W

F = W/2 = 50 N/2

F = 25 N

The force exerted by each string is 25 N

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A space probe is directly between two moons of a planet. If it is twice as far from moon A as it is from moon B, but the net for

Dennis_Churaev [7]

Answer:

c. Moon A is four times as massive as moon B

Explanation:

Let's assume the:

mass of the object = $m\,kilogram$

mass of the moon A = $M_A\,kilogram$

mass of the moon B = $M_B\,kilogram$

distance between the center of masses of the object and moon B = $r\,meters$

According to the given condition the object is twice as far from moon A as it is from moon B

∴distance between the center of masses of the object and moon B = $2r\,meters$

As we know, gravitational force of attraction is given by:

$F=G\frac{m_1.m_2}{r^2}$

According to the condition

Force on m due to $M_B=$ Force on m due to $M_A$

$G\frac{m.M_A}{(2r)^2} =G\frac{m.M_B}{(r)^2}$

$\frac{M_A}{4r^2} =\frac{M_B}{r^2}$

$M_A=4M_B$

3 0

2 years ago

Two bicycle tires are set rolling with the same initial speed of 3.30 m/s along a long, straight road, and the distance each tra

vredina [299]

Answer:

At low pressure- $\mu_{k}=0.02315$

At high pressure- $\mu_{k}=0.00445$

Explanation:

Initial speed, $V_{i}=3.3 m/s$

Final speed, $V_{f}=3.3/2= 1.65 m/s$

Net horizontal force due to rolling friction $F_{net}=\mu_{k}$ mg where m is mass, g is acceleration due to gravity, $\mu_{k}$ is coefficient of rolling friction