**Answer:**

12C

**Explanation:**

q = It

where q is the charge in coloumbs (c)

I is the current in amperes (A)

t is the time in seconds (s)

q = 0.2× 60 (1 minute is 60seconds)

q= 12C

What voltage was needed for 2 amps to flow through a 600 lamp?

**Answer:**

I don't kno..................................................w

A 320.-kg reindeer stands in the middle of the railroad tracks, frozen by the lights of an oncoming 10,000.-kg train that is traveling at 10.0 m/s. The engineer sees the reindeer but is unable to stop the train in time and the reindeer rides down the track sitting on the cowcatcher. What is the new combined velocity of the reindeer and train?

**Answer:**

**9.69 m/s**

**Explanation**

By the conservation of momentum, we can write the following equation:

[tex]\begin{gathered} p_i=p_f \\ m_1v_{i1}+m_2v_{i2}=(m_1+m_2)v_f \end{gathered}[/tex]Where m1 is the mass of the reindeer, m2 is the mass of the train, vi1 is the initial velocity of the reindeer, vi2 is the initial velocity of the train and vf is the final velocity of the reindeer and train.

So, replacing m1 = 320 kg, m2 = 10,000 kg, vi1 = 0 m/s and vi2 = 10 m/s, we get

[tex]\begin{gathered} 320(0)+10,000(10)=(320+10000)v_f_{} \\ 0+100,000=(10320)v_f \\ 100,000=10,320v_f \end{gathered}[/tex]Now, solving for the final velocity, we get

[tex]\begin{gathered} v_f=\frac{100,000}{10,320} \\ v_f=9.69\text{ m/s} \end{gathered}[/tex]Therefore, the new combined velocity of the reindeer and train is 9.69 m/s

A string 0.50 m long is stretched under a tension of 2.0 x 102 N and its fundamental frequency is 400 Hz. If the length if the string is shortened to 0.35 m and the tension is increased to 4.0 x 102 N, what is the new fundamental frequency?

Given,

The initial length of the string, L₁=0.50 m

The tension on the string, T₁=2.0×10² N

The initial fundamental frequency of the string, f₁=400 Hz

The length of the string after it was shortened, L₂=0.35 m

The increased tension on the string, T₂=4.0×10² N

The fundamental frequency of the string before it was shortened is given by,

[tex]f_1=\frac{\sqrt[]{\frac{T_1}{\mu}}}{2L_1}[/tex]Where μ is the mass per unit length of the string.

On rearranging the above equation,

[tex]\begin{gathered} 4L^2_1f^2_1=\frac{T_1}{\mu} \\ \Rightarrow\mu=\frac{T_1}{4L^2_1f^2_1} \end{gathered}[/tex]On substituting the known values,

[tex]\begin{gathered} \mu=\frac{2.0\times10^2}{4\times0.50^2\times400^2} \\ =1.25\times10^{-3}\text{ kg/m} \end{gathered}[/tex]The fundamental frequency after the string was shortened is given by,

[tex]f_2=\frac{\sqrt[]{\frac{T_2}{\mu}}}{2L_2}[/tex]On substituting the known values,

[tex]\begin{gathered} f_2=\frac{\sqrt[]{\frac{4\times10^2}{1.25\times10^{-3}}}}{2\times0.35} \\ =808.1\text{ Hz} \end{gathered}[/tex]**Thus the fundamental frequency after the string is shortened and the tension is increased is 808.1 Hz**

Which of the following statements correctly explains the relationship between FM and AM radio waves?

A.

FM radio waves carry less energy than AM radio waves. This is supported by the fact that FM radio waves have a shorter wavelength compared to AM radio waves.

B.

FM radio waves carry less energy than AM radio waves. This is supported by the fact that FM radio waves have a longer wavelength compared to AM radio waves.

C.

FM radio waves carry more energy than AM radio waves. This is supported by the fact that FM radio waves have a shorter wavelength compared to AM radio waves.

D.

FM radio waves carry more energy than AM radio waves. This is supported by the fact that FM radio waves have a longer wavelength

The relationship between the **AM **and the **FM **waves is that the wavelength of the FM waves is **longer **than that of the AM waves.

The** radio waves** are one of the waves that are part of the **electromagnetic spectrum** and they are used in the process of communication . The reason why the radio waves could be used for the process of communication is that the radio waves do have a long wavelength this they can travel for very long miles without a significant decrease in intensity of the waves.

Thus, **FM radio waves **carry less energy than** AM radio waves.** This is supported by the fact that FM radio waves have a longer wavelength compared to AM radio waves.

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Hello! So I need some help with this question. I don’t quite understand it and I did have another tutor figure it out and the answer was wrong. It’s not his answer of 58.04. Can you help me?

**ANSWER:**

**58 m/s**

**STEP-BY-STEP EXPLANATION:**

Given:

Initial velocity (u) = 13 m/s

The correct answer is 58 m/sDistance (d) = 400 m

Acceleration (a) = 4 m/s²

We apply the following formula to determine the final velocity:

[tex]\begin{gathered} v^2=u^2+2as \\ \\ \text{ We replacing:} \\ \\ v^2=13^2+2(4)(400) \\ \\ v^2=169+3200 \\ \\ v=\sqrt{3369} \\ \\ v=58.04\cong58\text{ m/s} \end{gathered}[/tex]**The correct answer is 58 m/s**

yo can anyone help me solve this rq i need it for tomorrow

a. 600 **meters**; b. 6 **minutes**; c. 7.5 minutes; d. 4 minutes; e. 5 minutes; f. 125m/min; g. 50m/min; h. 60m/min; i. 150m/min; j. 100m/min; k. 80m/min and l. 1m/s, 2.5m/s, 1.66m/s and 1.33m/s.

As per the** graph,**

The **motion **of Rachel and ben starts from t=0 and ends at t = 13minutes after covering **distance **of 600 meters.

a. As there motion is ending at 600 meters. They travel 600 meters to go to school.

b. Ben travels for 4 minutes first, after that he stops and then again starts from 8 minutes to 10 minutes. So, it takes 6 minutes for him to go to school.

c. Rachel travels for 5 minutes first, after that she stops and then again starts from 10 minutes to 12.5 minutes. So, it takes 7.5 minutes for her to go to school.

d. Ben stops from 4 to 8 minutes on his way to school. After that he stops only at school so it would not be counted. So he stops for 4 minutes.

e. Rachel stop from 5 minutes to 10 minutes on her way. So, she stops for 5 minutes in her way to school.

f. Ben's speed in first half,

**Speed = distance in first half/time taken in first half.**

Ben's Speed in first half = 500/4

Ben's Speed in first half = 125 m/min.

g. Ben's speed in last part,

Speed = distance in last part/time taken in last part.

Ben's Speed in last part = 100/2

Ben's Speed in last part = 50m/min.

h. Rachel's speed in first half,

Speed = distance in first half/time taken in first half.

Rachel's Speed in first half = 300/5

Rachel's Speed in first half = 60 m/min.

i. Rachel's speed in last part,

Speed = distance in last part/time taken in last part

Rachel's Speed in last part = 300/2

Rachel's Speed in last part = 150m/min.

j. Ben's average speed = total distance/total time

Total distance = 600

Total time = 6 minutes

Average speed of Ben = 600/6 = 100 m/min.

k. Rachel's average speed = total distance/total time

Total distance = 600

Total time = 7.5 minutes

Average speed of Rachel = 600/7.5 = 80m/min.

l. **Converting m/min** to **m/s **by **dividing **by 60.

Converting speed in h = 60/60 = 1m/s.

Converting speed in i = 150/60 = 2.5m/s.

Converting speed in j = 100/60 = 1.66m/s.

Converting speed in k = 80/60 = 1.33m/s.

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Which of the following numbers has the greatest number of significant figures?a.) 144.5b.) 0.009514 c.) 10.507d.) 6.948 ✕ 10

So, the number with the greatest number of significant figures is:

0.009514. **[Option B]**

A block of mass M is hung by ropes as shown. The system is in

equilibrium. The point O represents the knot, the junction of the

three ropes. The angle theta is given to be 30o

. Which of the

following statements is true concerning the magnitudes of the three

forces in equilibrium?

Their **magnitude **of force must be the same for the **tensions** to equal out.

An object at **equilibrium** has zero **acceleration** so both the magnitude and direction of the object's velocity must be constant. When the body is in equilibrium the forces are balanced. Balanced is a keyword used to describe a balanced state. Therefore the net force is zero and the acceleration is 0 m/s/s. The acceleration of a body in equilibrium must be 0 m/s/s

For a body to be in **equilibrium** it must not be** accelerating**. This means that both the net force and net torque on the object must be zero. Clearly when in equilibrium, the net force on the object is zero. According to Newton's second law of motion if the net force is zero the acceleration is also zero. When the acceleration is zero the velocity and thus the velocity is constant by definition. Option d does not apply to equilibrium in this case.

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Starting from rest, a car accelerates at 2.0 m/s2 up a hill that is inclined 5.5° above the horizontal. How far (a) horizontally and (b) vertically has the car traveled in 12 s?

**Given data:**

* The acceleration of the car is,

[tex]a=2ms^{-2}^{}[/tex]* The angle of the inclined plane is 5.5 degree.

* The time taken by the car is 12 s.

* The initial velocity of the car is 0 m/s.

**Solution:**

By the kinematics equation, the final velocity of the car on the inclined plane is,

[tex]v-u=at[/tex]where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the tiem taken,

Subsituting the known values,

[tex]\begin{gathered} v-0=2\times12 \\ v=24\text{ m/s} \end{gathered}[/tex]By the kinematics equation, the distance tarveled by the car on the inclined plane is,

[tex]v^2-u^2=2aS[/tex]where S is the distance tarveled on the inclined plane,

Substituting the known values,

[tex]\begin{gathered} 24^2-0=2\times2\times S \\ 576=4\times S \\ S=\frac{576}{4} \\ S=144\text{ m} \end{gathered}[/tex]The diagrammatic representation of the car on the inclined plane is,

The distance traveled by the car in the horizontal direction or along x-axis is,

[tex]\begin{gathered} S_x=S\cos (5.5^{\circ}) \\ S_x=144\times\cos (5.5^{\circ}) \\ S_x=143.34\text{ m} \end{gathered}[/tex]**Thus, the distance tarveled by the car in the horizontal direction is 143.34 meter.**

The distance tarveled by the car in the vertical direction or y-axis is,

[tex]\begin{gathered} S_y=S\sin (5.5^{\circ}) \\ S_y=144\times\sin (5.5^{\circ}) \\ S_y=13.8\text{ m} \end{gathered}[/tex]**Thus, the distance tarveled by the car in the vertical direction is 13.8 meter.**

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