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Anuta_ua [19.1K]

3 years ago

7

An apparatus like the one Cavendish used to find G has large lead balls that are 8.4 kg in mass and small ones that are 0.061 kg

. The center of a large ball is separated by 0.057 m from the center of a small ball. Find the magnitude of the gravitational force between the masses if the value of the universal gravitational constant is 6.67259 × 10−11 Nm2/kg2

1 answer:

garik1379 [7]3 years ago

6 0

Answer:

The gravitational force is $1.05\times10^{-8}\ N$

Explanation:

Given that,

Mass of large ball = 8.4 kg

Mass of small ball = 0.061 kg

Separation = 0.057 m

Gravitational constant $G= 6.67\times10^{-11}\ Nm^2/kg^2$

We need to calculate the gravitational force

Using formula of gravitational force

$F= \dfrac{Gm_{1}m_{2}}{r^2}$

Put the value into the formula

$F=\dfrac{6.67259\times10^{-11}\times8.4\times0.061}{(0.057)^2}$

$F=1.05\times10^{-8}\ N$

Hence, The gravitational force is $1.05\times10^{-8}\ N$

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A spinning disk is rotating at a rate of 5 rad/s in the positive counterclock-wise direction. If the disk is subjected to an ang

Anit [1.1K]

Answer:

ωf = 13 rad/s

Explanation:

The angular acceleration, by definition, is just the rate of change of the angular velocity with respect to time, as follows:
α = Δω/Δt = (ωf-ω₀) / (tfi-t₀)
Choosing t₀ = 0, and rearranging terms, we have

$\omega_{f} = \omega_{o} + \alpha *t (1)$

where ω₀ = 5 rad/s, t = 4 s, α = 2 rad/s2

Replacing these values in (1) and solving for ωf, we get:

$\omega_{f} = 5 rad/s + (2 rad/s2*4 s) = 13 rad/s (2)$

The wheel's angular velocity after 4s is 13 rad/s.

3 0

3 years ago

A vertical straight wire carrying an upward 28-A current exerts an attractive force per unit length of 7.83 X 10 N/m on a second

JulijaS [17]

Answer:

$i_2 = 978750 A$

Since the force between wires is attraction type of force so current must be flowing in upward direction

Explanation:

Force per unit length between two current carrying wires is given by the formula

$F = \frac{\mu_0 i_1 i_2}{2 \pi d}$

here we know that

$F = 7.83 \times 10 N/m$

$d = 7.0 cm = 0.07 m$

$i_1 = 28 A$

now we will have

$F = \frac{4\pi \times 10^{-7} (28.0)(i_2)}{2\pi (0.07)}$

$7.83 \times 10 = \frac{2\times 10^{-7} (28 A)(i_2)}{0.07}$

$i_2 = 978750 A$

Since the force between wires is attraction type of force so current must be flowing in upward direction

3 0

3 years ago

Even when the head is held erect, as in the figure below, its center of mass is not directly over the principal point of support

alexandr1967 [171]

We are asked to determine the force required by the neck muscle in order to keep the head in equilibrium. To do that we will add the torques produced by the muscle force and the weight of the head. We will use torque in the clockwise direction to be negative, therefore, we have:

$\Sigma T=r_{M\perp}(F_M)-r_{W\perp}(W)$

Since we want to determine the forces when the system is at equilibrium this means that the total sum of torque is zero:

$r_{M\perp}(F_M)-r_{W\perp}(W)=0$

Now, we solve for the force of the muscle. First, we add the torque of the weight to both sides:

$r_{M\perp}(F_M)=r_{W\perp}(W)$

Now, we divide by the distance of the muscle:

$(F_M)=\frac{r_{W\perp}(W)}{r_{M\perp}}$

Now, we substitute the values:

$F_M=\frac{(2.4cm)(50N)}{5.1cm}$

Now, we solve the operations:

$F_M=23.53N$

Therefore, the force exerted by the muscles is 23.53 Newtons.

Part B. To determine the force on the pivot we will add the forces we add the vertical forces:

$\Sigma F_v=F_j-F_M-W$

Since there is no vertical movement the sum of vertical forces is zero:

$F_j-F_M-W=0$

Now, we add the force of the muscle and the weight to both sides to solve for the force on the pivot:

$F_j=F_M+W$

Now, we plug in the values:

$F_j=23.53N+50N$

Solving the operations:

$F_j=73.53N$

Therefore, the force is 73.53 Newtons.

8 0

1 year ago

Which sentence states Newton's second law?

Shalnov [3]

Answer:

Force is equal to the change in momentum per change in time.

Explanation:

That situation is described by Newton's Second Law of Motion. According to NASA, this law states, "Force is equal to the change in momentum per change in time. For a constant mass, force equals mass times acceleration." This is written in mathematical form as Force = mass.

6 0

2 years ago

A force of 48 newtons is required to start a 5.0 kg box moving across a horizontal concrete floor. What is the coefficient of st

Soloha48 [4]

1) The coefficient of static friction is 0.980

2) The coefficient of kinetic friction is 0.908

Explanation:

1)

In the first situation, the box is still at rest. There are two forces acting on the box:

- The force of push, F, forward

- The force of static friction, $F_f$

Since the box is in equilibrium we have

$F_f = F$ (1)

The value of frictional force changes from zero to a maximum value which is given by:

$F_f = \mu_s mg$ (2)

where

$\mu_s$ is the coefficient of static friction

m is the mass of the box

g is the acceleration of gravity

So, if the force F needed to put the box in motion is 48 N, it means that this is also the maximum value of the force of friction. So, we can combine eq.(1) and (2) to find the coefficient of static friction:

$\mu_s = \frac{F}{mg}$

where:

F = 48 N

m = 5.0 kg

$g=9.8 m/s^2$

Substituting,

$\mu_s = \frac{48}{(5.0)(9.8)}=0.980$

2)

In this second situation, the object is already in motion, so the equation of motion is:

$F-F_f = ma$

where

F = 48 N is the force applied forward

$a = 0.70 m/s^2$ is the acceleration of the box

$F_f = \mu_k mg$ is the force of kinetic friction, where

$\mu_k$ is the coefficient of kinetic friction

We can therefore rearrange the equation to find the coefficient:

$F-\mu_k mg = ma\\\mu_k mg = F-ma\\\mu_k = \frac{F-ma}{mg}=\frac{48-(5.0)(0.70)}{(5.0)(9.8)}=0.908$

Learn more about friction:

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7 0

3 years ago