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a_sh-v [17]
2 years ago
15

A speedboat is moving at a rate of 45 km per hour travels a distance of 27 km. How long did it take to go to 27 km.

Physics
1 answer:
8090 [49]2 years ago
8 0

Answer:

I think it is 1.67

Explanation:

I multiply the 45 and 27 cause it says rate of which means to multiply.

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Objects in space are moving at a constant velocity in a straight line.
tigry1 [53]

Answer:

C

Explanation:

An object in motion will stay in motion unless acted on by a net positive or negative force.

For answer A. If the object were to be in an orbit, it would inevitably accelerate due to it being acted on by the gravitational force from the object it is orbiting. At different points in the orbit, the object will move at different speeds and continuously transfer between kinetic and potential energy.

For answer B. The object would would not stop their motion. In order for the object to lose energy, it would have to transfer it through friction or through its interaction with a gravitational field.

For answer D. No energy is "required" to maintain constant motion unless the object is willingly fighting against a resistive force like friction or a graviational well.

8 0
2 years ago
A +2.00nc point charge is at the origin, and a second -5.00nc point charge is on the x-axis at x = 0.800m find the magnitude of
Damm [24]
You have to reduce 2.00 an5.00 I order to use the×that=0.800
3 0
3 years ago
One negative change you may encounter as a student or an employee​
matrenka [14]

Answer:

you may get bullied or teased for being a differrent race, ethnic.

7 0
3 years ago
Read 2 more answers
A baseball pitcher throws a ball horizontally at a speed of 34.0 m/s. A catcher is 18.6 m away from the pitcher. Find the magnit
Sidana [21]

To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.

The trajectory equation from the motion kinematic equations is given by

y = \frac{1}{2} at^2+v_0t+y_0

Where,

a = acceleration

t = time

v_0 = Initial velocity

y_0 = initial position

In addition to this we know that speed, speed is the change of position in relation to time. So

v = \frac{x}{t}

x = Displacement

t = time

With the data we have we can find the time as well

v = \frac{x}{t}

t = \frac{x}{v}

t = \frac{18.6}{34}

t = 0.547s

With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,

y = \frac{1}{2} at^2+v_0t+y_0

y = \frac{1}{2} gt^2+0+0

y = \frac{1}{2} gt^2

y = \frac{1}{2} 9.8*0.547^2

y = 1.46m

Therefore the vertical distance that the ball drops as it moves from the pitcher to the catcher is 1.46m.

6 0
3 years ago
Raindrops fall vertically at 7.5 m/s relative to the Earth. What does an observer in a car moving at 20.2 m/s in a straight line
Vilka [71]

Answer:

vDP = 21.7454 m/s

θ = 200.3693°

Explanation:

Given

vDE = 7.5 m/s

vPE = 20.2 m/s

Required:  vDP

Assume that

vDE to be in direction of - j

vPE to be in direction of i

According to relative motion concept the velocity vDP is given by

vDP = vDE - vPE     (I)

Substitute in (I) to get that

vDP = - 7.5 j - 20.2 i

The magnitude of vDP is given by

vDP = √((- 7.5)²+(- 20.2)²) m/s =  21.7454 m/s

θ = Arctan (- 7.5/- 20.2) = 20.3693°

θ is in 3rd quadrant so add 180°

θ = 20.3693° + 180° = 200.3693°

4 0
3 years ago
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