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2 years ago

If a bowl of fruit is sitting on a table in a state of rest, what is the net force acting on it?

2 answers:

8 0

A balanced force is upon it because balanced means both sides are equal so they cancel each other out, while unbalanced means one side is weaker/stronger than the other and in this case its in a state of rest so the net force would be equal/balanced.

Anna11 [10]2 years ago

7 0

0 N is the correct answer. No force is acting on it.

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A uniform 40-N board supports two children weighing 500 N and 350 N. If the support is at the center of the board and the 500-N

serg [7]

Answer:

b= 2.14 m

Explanation:

Given that

Weight of the board ,wt = 40 N

Wight of the first children , wt₁=500 N

Weight of the second children ,wt₂ = 350 N

The distance of the 500 N child from center ,a= 1.5 m

lets take distance of the 350 N child from center = b m

Now by taking the moment about the center of the board

We know that moment = Force x Perpendicular distance from the force

wt₁ x a = wt₂ x b

500 x 1.5 = 350 x b

b= 2.14 m

Therefore the distance of the 350 N weight child from the center is 2.14 m.

5 0

1 year ago

On a snowy day, when the coefficient of friction μs between a car’s tires and the road is 0.50, the maximum speed that the car c

Nana76 [90]

Answer:

28.1 mph

Explanation:

The force of friction acting on the car provides the centripetal force that keeps the car in circular motion around the curve, so we can write:

$F=\mu mg = m \frac{v^2}{r}$ (1)

where

$\mu$ is the coefficient of friction

m is the mass of the car

g = 9.8 m/s^2 is the acceleration due to gravity

v is the maximum speed of the car

r is the radius of the trajectory

On the snowy day,

$\mu=0.50\\v = 20 mph = 8.9 m/s$

So the radius of the curve is

$r=\frac{v^2}{\mu g}=\frac{(8.9)^2}{(0.50)(9.8)}=16.1 m$

Now we can use this value and re-arrange again the eq. (1) to find the maximum speed of the car on a sunny day, when $\mu=1.0$ . We find:

$v=\sqrt{\mu g r}=\sqrt{(1.0)(9.8)(8.9)}=12.6 m/s=28.1 mph$

7 0

2 years ago

Read 2 more answers

Pulling out of a dive, the pilot of an airplane guides his plane into a vertical circle with a radius of 600 m. At the bottom of

adoni [48]

Answer:

3311N

Explanation:

r = radius = 600m

V = speed = 150m/s

Mass = weight = 70kg

The weight of pilot when calculated due to circular motion

W = tv

Fv = mv²/r

Fv = 70x150²/600

Fv = 79x22500/600

= 15750000/600

= 2625N

Real Weight of the pilot = m x g

= 70 x 9.8

= 686N

The apparent Weight is calculated by

Mv²/r + mg

= 2625N + 686N

= 3311 N

Therefore the apparent Weight is 3311N

6 0

1 year ago

The amount of light that undergoes reflection or transmission is demonstrated by how bright the reflected or transmitted ray is.

melamori03 [73]

If you subscribe I’ll answer QF Aotrx

8 0

2 years ago

An air-track cart with mass m1=0.28kg and initial speed v0=0.75m/s collides with and sticks to a second cart that is at rest ini

arsen [322]

Kinetic energy is calculated through the equation,

KE = 0.5mv²

At initial conditions,

m₁: KE = 0.5(0.28 kg)(0.75 m/s)² = 0.07875 J

m₂ : KE = 0.5(0.45 kg)(0 m/s)² = 0 J

Due to the momentum balance,

m₁v₁ + m₂v₂ = (m₁ + m₂)(V)

Substituting the known values,

(0.29 kg)(0.75 m/s) + (0.43 kg)(0 m/s) = (0.28 kg + 0.43 kg)(V)

V = 0.2977 m/s

The kinetic energy is,
KE = (0.5)(0.28 kg + 0.43 kg)(0.2977 m/s)²
KE = 0.03146 J

The difference between the kinetic energies is 0.0473 J.

7 0

2 years ago