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2 years ago

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The acceleration due to gravity for any object, including 1 washer on the string, is always assumed to be m/s2. The mass of 3 wa

shers, when converted to kg, is kg. The applied force of 3 washers will increase the applied force on the car to N.

2 answers:

Andrews [41]2 years ago

6 0

Answer: 10, 0.0147, 0.147

Explanation: Correct on edge2020

Pachacha [2.7K]2 years ago

5 0

Answer:

The force will increase in proportion to the mass of the objects

Explanation:

The acceleration due to gravity is always the same. It is expressed in meters per second squared or m/s². The figure of 9.81 m/s² is an average value that was taken after calculating the acceleration under different surfaces. In fact, the acceleration differs depending on the shape of the part of the earth in relation to the earth's magnetic field and force.

Thus, if one washer was 20 kg, the acceleration being 9.81 m/s² the weight will be:

F = ma

= $20 * 9.81\\= 196.2 N$

If there are there washers, the weight will be:

F = 3 * 20 * 9.81

= 588.6 N

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A uniform sphere with mass M and radius R is rotating with angular speed ω1 about a frictionless axle along a diameter of the sp

liq [111]

Answer:

$W_2=\sqrt{\frac{3}{5} }W_1$

Explanation:

For the first ball, the moment of inertia and the kinetic energy is:

$I_1 =\frac{2}{5}MR^2$

$K_1 = \frac{1}{2}IW_1^2$

So, replacing, we get that:

$K_1 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2$

At the same way, the moment of inertia and kinetic energy for second ball is:

$I_2 =\frac{2}{3}MR^2$

$K_2 = \frac{1}{2}IW_2^2$

So:

$K_2 = \frac{1}{2}(\frac{2}{3}MR^2)W_2^2$

Then, $K_2$ is equal to $K_1$ , so:

$K_2 = K_1$

$\frac{1}{2}(\frac{2}{3}MR^2)W_2^2 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2$

$\frac{1}{3}MR^2W_2^2 = \frac{1}{5}MR^2W_1^2$

$\frac{1}{3}W_2^2 = \frac{1}{5}W_1^2$

Finally, solving for $W_2$ , we get:

$W_2=\sqrt{\frac{3}{5} }W_1$

5 0

2 years ago

In certain cases, using both the momentum principle and energy principle to analyze a system is useful, as they each can reveal

SpyIntel [72]

Answer:

A) F_g = 26284.48 N

B) v = 7404.18 m/s

C) E = 19.19 × 10^(10) J

Explanation:

We are given;

Mass of satellite; m = 3500 kg

Mass of the earth; M = 6 x 10²⁴ Kg

Earth circular orbit radius; R = 7.3 x 10⁶ m

A) Formula for the gravitational force is;

F_g = GmM/r²

Where G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Plugging in the relevant values, we have;

F_g = (6.67 × 10^(-11) × 3500 × 6 x 10²⁴)/(7.3 x 10⁶)²

F_g = 26284.48 N

B) From the momentum principle, we have that the gravitational force is equal to the centripetal force.

Thus;

GmM/r² = mv²/r

Making v th subject, we have;

v = √(GM/r)

Plugging in the relevant values;

v = √(6.67 × 10^(-11) × 6 x 10²⁴)/(7.3 x 10⁶))

v = 7404.18 m/s

C) From the energy principle, the minimum amount of work is given by;

E = (GmM/r) - ½mv²

Plugging in the relevant values;

E = [(6.67 × 10^(-11) × 3500 × 6 × 10²⁴)/(7.3 x 10⁶)] - (½ × 3500 × 7404.18)

E = 19.19 × 10^(10) J

5 0

2 years ago

a partially inflated weather balloon has a volume of 1.56 * 10^3 L and a pressure of 98.9 lPa. What is the volume of the balloon

e-lub [12.9K]

To solve the problem, we enumerate all the given first. Then the required and lastly the solution.

Given:

V1= 1.56x10^3 L = 1560 L P2 = 44.1 kPa
P1 = 98.9 kPa

Required: V2

Solution:

Assuming the gas is ideal. Ideal gas follows Boyle's Law which states that at a given temperature the product of pressure and volume of a gas is constant. In equation,

PV = k

Applying to the problem, we have

P1*V1 = P2*V2
(98.9 kPa)*(1560 L) = (44.1 kPa)*V2
V2 = 3498.5 L

<em>ANSWER: V2 = 3498.5 L</em>

7 0

2 years ago

A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±qseparated by a distance s. T

atroni [7]

Answer:

The magnitude of the force on the dipole due to the charge Q = $\rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.$

The magnitude of the torque on the dipole = $\rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi}\dfrac{2qQs^2}{r^3}.$

Explanation:

Given that a point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q, separated by a distance s and the charge Q is located in the plane that bisects the dipole.

The magnitude of the electric field that the dipole exerts at the position where the charge Q is held is given by

$\rm E = \dfrac{k2qs}{(r^2+s^2)^{3/2}}.$

<em>where</em>,

k is the Coulomb's constant, having value = $\dfrac{1}{4\pi \epsilon_o}$

$\epsilon_o$ is the electrical permittivity of free space.

Also, r>>s, therefore, $\rm r^2+s^2\approx r^2.$

$\rm E = \dfrac{k2qs}{(r^2)^{3/2}}=\dfrac{k2qs}{r^3}.$

The magnitude of the electric force F on a charge q placed at a point and the magnitude of the electric field E at that point are related as

$\rm F=qE$

Therefore, the electric force on the charge Q due to the dipole is given by

$\rm F=Q\dfrac{k2qs}{r^3}=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}.$

According to Newton's third law of motion, the magnitude of the force exerted by the dipole on the charge Q is same as the magnitude of the force exerted by the charge on the dipole.

Thus, the magnitude of the force on the dipole due to the charge Q = $\dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.$

The magnitude of the torque on the dipole is given by

$\rm \tau = Fs\ \sin\theta$

Since the charge Q is placed in the plane that bisects the dipole, therefore, $\theta = 90^\circ$ .

$\rm \tau = \dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}\cdot s\cdot 1=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs^2}{r^3}.$

4 0

2 years ago

A string is wrapped around a pulley with a radius of 2.0 cm and no appreciable friction in its axle. The pulley is initially not

stiks02 [169]

Answer:

2

Explanation:

2

5 0

2 years ago

Read 2 more answers