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tatyana61 [14]
3 years ago
15

A small 25 kilogram canoe is floating downriver at a speed of 4 m/s. What is the canoe's kinetic energy?

Physics
1 answer:
Marysya12 [62]3 years ago
8 0
The canoe's KE is 200 'cause you have to multiply the mass to the squared(times itself) velocity and once you got the product, divide it by 2
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A conductor carrying a current I = 13.8 A is directed along the positive x axis and perpendicular to a uniform magnetic field. A
Scorpion4ik [409]

Answer:

a) B=0.008 T

b) +z direction

Explanation:

<u>solution:</u>

a) The magnetic force:

        F=i*l*B

Solve for B:  

      B=0.008 T

b) According to the left hand rule, the magnetic field is in the +z direction

3 0
3 years ago
While driving fast around a sharp right turn, you find yourself pressing against the car door. What is happening?
sergiy2304 [10]

Answer:

option C

Explanation:

The correct answer is option C

When the driver takes the sharp right turn the door will exert rightward pressure on the driver.

When the driver takes the sudden right turn the tendency of the body is to be in the straight line by the vehicle moves in the circular path so, as the vehicle turns it applies a rightward force on you.

The pushing of the door to you because of the centripetal force acting on the car due to sudden sharp turn.

3 0
3 years ago
A center-seeking force related to acceleration is _______ force.
Gennadij [26K]
<span>A center-seeking force related to acceleration is centripetal force. The answer is letter A. The rest of the choices do not answer the question above.</span>
4 0
3 years ago
When you run outside you transform what energy into what energy
erica [24]
Potential energy and kinetic energy
6 0
4 years ago
A small propeller airplane can comfortably achieve a high enough speed to take off on a runway that is 1/4 mile long. A large, f
QveST [7]

Answer:

1 mile

Explanation:

We can use the following equation of motion to solve for this problem:

v^2 - v_0^2 = 2a\Delta s

where v m/s is the final take-off velocity of the airplane, v_0 = 0 initial velocity of the can when it starts from rest, a is the acceleration of the airplanes, which are the same, and \Delta s is the distance traveled before takeoff, which is minimum runway length:

v^2 - 0^2 = 2a\Delta s

\Delta s = \frac{v^2}{2a}

From here we can calculate the distance ratio

\frac{\Delta s_1}{\Delta s_2} = \frac{v_1^2/2a_1}{v_2^2/2a_2}

\frac{\Delta s_1}{\Delta s_2} = \left(\frac{v_1}{v_2}\right)^2\frac{a_2}{a_1}

Since the 2nd airplane has the same acceleration but twice the velocity

\frac{\Delta s_1}{\Delta s_2} = 0.5^2* 1

\Delta s_2 = 4 \Delta s_1 = 4*(1/4) = 1 mile

So the minimum runway length is 1 mile

6 0
3 years ago
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