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2 years ago

13. An airplane travels at a constant speed of 800 kilometers per hour (km/h).

Physics

1 answer:

kramer2 years ago

5 0

Answer:

The time taken for the airplane to cover the distance is 15 minutes.

Explanation:

Given;

velocity of airplane, u = 800 km/h

distance traveled by the airplane, s = 200 km

The time of this motion is given by;

s = ut + ¹/₂at²

at constant velocity, acceleration, a = 0

s = ut

t = s / u

t = (200 km) / (800 km/h)

t = 0.25 hour = 15 mins

Therefore, the time taken for the airplane to cover the distance is 15 minutes.

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A 5.20-N force is applied to a 1.05-kg object to accelerate it rightwards. The object encounters 3.29-N of friction. Determine t

tatiyna

Answer:

Explanation:

We will use the equation F - f = ma, which is just a fancy way of stating Newton's 2nd Law. For us:

F = 5.20 to the right (+)

f = 3.29 to the left (-)

m = 1.05 kg. Therefore,

5.20 - 3.29 = 1.05a and

1.91 = 1.05a so

a = 1.82 m/s/s to the right

8 0

2 years ago

The planet whose distance from the sun equals 1 <br> a.u. is: venus earth mars jupiter

son4ous [18]

The answer is Earth. hope this helped

7 0

2 years ago

Read 2 more answers

A diamond is cut such that the angle between its top surface and its bottom surface is α. For α=45∘, find the largest possible v

Reil [10]

Answer:

i = 61 degree

Explanation:

Given,

$\alpha =45^{o}$

Now, by the snell's law

$sin\theta_c=\frac{1}{n}\\sin\theta_c=\frac{1}{2.450}\\\theta_c=24.09^o$

Now,

Sin i / sin r = n 2 / n 1

sin i / sin r (45 - 24.09) = 2.45 / 1

i = 60.97 degree

4 0

2 years ago

Pls help in these 2 questions

Nadya [2.5K]

Answer:

Taking forces along the plane

F cos θ - M g sin θ -100 = M a net of forces along the plane

F = (M a + M g * .5 + 100) / .866 solving for F

F = (80 * 1.5 + 80 * 9.8 * .5 + 100) / .866 = 707 N

F = 707 N acting along the plane

Fn = F sin θ + M g cos θ forces acting perpendicular to plane

Fn = 707 * 1/2 + 80 * 9.8 * .866 = 1030 Newtons forces normal to plane

(this would give a coefficient of friction of 100 / 1030 = .097 = Fn)

4 0

1 year ago

A ball with 100 J of PE is released from a height of 10 m. What will be the KE of the ball at 5

harkovskaia [24]

Answer:

The kinetic energy is: 50[J]

Explanation:

The ball is having a potential energy of 100 [J], therefore

PE = [J]

The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.

$E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\$

In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.

When the elevation is 5 [m], we have a potential energy of

$P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\$

This energy is equal to the kinetic energy, therefore

Ke= 50 [J]

8 0

2 years ago