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3 years ago

Brant pushes a toy car with a force of 50 N. The force of friction as the car moves is 5 N. What is the net force on the car?

Physics

1 answer:

ICE Princess25 [194]3 years ago

4 0

net force= the sum of all the forces acting on the object

the frictional force is opposite to the direction of motion of the object

If the object moves to the right (positive) then the friction force moves to the left (negative).

so the net force: 50 N - 5 N = 45 N (B)

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Public television station KQED in San Francisco broadcasts a sinusoidal radio signal at a power of 777 kW. Assume that the wave

ehidna [41]

Answer:

A) P = 3.3 × 10^(-11) Pa

B) Amplitude of electric field = 1.931 N/C

Amplitude of magnetic field = 6.44 × 10^(-9) T

C) μ_av = 1.65 × 10^(-11) J/m³

D) 50% each for the electric and magnetic field

Explanation:

A) First of all let's calculate intensity.

I = P_av/A

We are given;

P_av = 777 KW = 777,000 W

Distance = 5 km = 5000 m

Thus;

I = 777000/(2π × 5000²)

I = 0.00495 W/m²

Now, the average pressure would be given by the formula;

P = 2I/C

Where C is speed of light = 3 × 10^(8) m/s

P = (2 × 0.00495)/(3 × 10^(8))

P = 3.3 × 10^(-11) Pa

B) Formula for the amplitude of the electric field is gotten from;

E_max = √[2I/(εo•c)].

Where εo is the Permittivity of free space with a constant value of 8.85 × 10^(−12) c²/N.mm²

I and c remain as before.

Thus;

E_max = √[(2 × 0.00495)/(8.85 × 10^(−12) × 3 × 10^(8))]

E_max = √3.72881355932

E_max = 1.931 N/C

Formula for amplitude of magnetic field is gotten from;

B_max = E_max/c

B_max = 1.931/(3 × 10^(8))

B_max = 6.44 × 10^(-9) T

C) Formula for average density is;

μ_av = εo(E_rms)²

Now, E_rms = E_max/√2

Thus;

E_rms = 1.931/√2

μ_av = 8.85 × 10^(−12) × (1.931/√2)²

μ_av = 1.65 × 10^(-11) J/m³

D) The energy density for the electric and magnetic field is the same. So both of them will have 50% of the energy density.

4 0

3 years ago

A car is traveling at 108 km/h, stuck behind a slower car. Finally the road is clear and the car pulls over to make a pass. The

mezya [45]

Answer:

The average acceleration of the car is 2.143 meters per square second.

Explanation:

Let assume that car accelerates uniformly, in that case, we can obtain the value of acceleration by using the following equation of motion:

$v = v_{o}+a\cdot t$

Where:

$v_{o}$ - Initial velocity, measured in meters per second.

$v$ - Final velocity, measured in meters per second.

$a$ - Acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

Now, we clear acceleration within expression:

$a = \frac{v-v_{o}}{t}$

Initial and final velocities are now converted from kilometers per hour into meters per second:

$v_{o} = \left(108\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s} \right)$

$v_{o} = 30\,\frac{m}{s}$

$v = \left(135\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s} \right)$

$v = 37.5\,\frac{m}{s}$

If we know that $t = 3.5\,s$ , then, the average acceleration of the car is:

$a = \frac{37.5\,\frac{m}{s}-30\,\frac{m}{s} }{3.5\,s}$

$a = 2.143\,\frac{m}{s^{2}}$

The average acceleration of the car is 2.143 meters per square second.

8 0

3 years ago

A dog runs with an initial speed of 1.5 m/s on a floor. It slides to a stop with an acceleration of -0.35 m/s^2. How long does i

Neko [114]

Answer:

3.214 m

Explanation:

Here object is moving in a constant acceleration. Then we can use motion equations to find the total distance

V² = U² + 2as

0 = 1.5² + 2×-0.35 × s

s = 3.214 m

symbols has usual meanings.

6 0

3 years ago

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We know that the earth's axis is tilted 23 ½ degrees. On or about June 21 or 22 each year, the summer solstice occurs for those

Dima020 [189]

<span>D) The sun's rays will never be directly overhead. The latitude of 23 ½ degrees north is known as the Tropic of Cancer. Above this imaginary line the sun's rays hit earth with decreased angles.</span>

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4 years ago

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In the equilibrium position, the 30-kg cylinder causes a static defl ection of 50 mm in the coiled spring. If the cylinder is de

MArishka [77]

Answer:

2.23 Hz

Explanation:

From the attached diagram below; there exists a diagrammatic representation of the equilibrium position of the cylinder.

The equilibrium position of the spring is expressed as:

mg = K $\delta _{st}$

where;

m = mass of the object

g = acceleration due to gravity

K = spring constant

$\delta _{st}$ = static deflection of the string

Given that:

m = 30 kg

g = 9.81 m/s²

$\delta _{st}$ = 50 mm = 50 × $\frac{1 \ m}{1000 \ m}$

= 0.05 m

Then;

$30 * 9.81= k * 0.05\\k = \frac{30*9.81}{0.05} \\k = 5886 N/m$

From here; let us find the angular velocity which will be needed to determine the natural frequency aftewards.

The angular velocity of the cylinder can be expressed by the formula:

$\omega_{n} = \sqrt{\frac{k}{m}}$

$\omega_{n} = \sqrt{\frac{5886}{30}}$

$\omega_{n} = \sqrt{196.2}$

$\omega_{n} = 14.007141 \ \ rad/s$

Finally; the natural frequency $f_n$ can be calculated by using the equation

$f_n = \frac{\omega_n}{2 \ \pi }$

$f_n = \frac{14.007141}{2 \ \pi }$

$f_n$ = 2.229305729

$f_n$ ≅ 2.23 Hz

Thus; the resulting natural frequency of the vertical vibration of the cylinder = 2.23 Hz

3 0

3 years ago