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Musya8 [376]
1 year ago
15

A 100-kg box is sitting on a 10 degree incline with a coefficient of friction of 0.5. At what angle must the incline be raised t

o start sliding on the box?
Physics
1 answer:
Marta_Voda [28]1 year ago
5 0

Answer:

26.56°

Explanation:

μ= 0.5

F = ?

N =?

Fs = μN

Fs = force acting due to friction (on an inclined plane)

N = normal force.

μ= coefficient of friction

μ= tanΘ

μ= tan 0.5

Θ = tan⁻¹ 0.5

Θ = 26.56°

The angle in which is required for the box to start moving across the inclined plane is 26.56°

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How much energy is needed to change the speed of a 1600 kg sport utility vehicle from 15.0 m/s to 40.0 m/s?
podryga [215]

Answer:

Energy needed = 1100 kJ

Explanation:

Energy needed = Change in kinetic energy

Initial velocity = 15 m/s

Mass, m = 1600 kg

\texttt{Initial kinetic energy = }\frac{1}{2}mv^2=0.5\times 1600\times 15^2=180000J

Final velocity = 40 m/s

\texttt{Final kinetic energy = }\frac{1}{2}mv^2=0.5\times 1600\times 40^2=1280000J

Energy needed = Change in kinetic energy = 1280000-180000 = 1100000J

Energy needed = 1100 kJ

4 0
2 years ago
A plane initially traveling at 200 m/s due west experiences a 10 m/s head wind coming from the opposite direction. A). What will
Hitman42 [59]
Ok so it would be late and the relative velocity would be 190 m/s because 200 m/s - 10 m/s is 190 m/s. Hope this helps.
8 0
2 years ago
A rescue helicopter wants to drop a package of supplies to isolated mountain climbers on a rocky ridge 200 m below. If the helic
andreev551 [17]

Answer:

a) 447.21m

b) -62.99 m/s

c)94.17 m/s

Explanation:

This situation we can divide in 2 parts:

⇒ Vertical : y =-200 m

y =1/2 at²

-200 = 1/2 *(-9.81)*t²

t= 6.388766 s

⇒Horizontal: Vx = Δx/Δt

Δx = 70 * 6.388766 = 447.21 m

b) ⇒ Horizontal

Vx = Δx/Δt ⇒ 70 = 400 /Δt

Δt= 5.7142857 s

⇒ Vertical:

y = v0t + 1/2 at²

-200 = v(5.7142857) + 1/2 *(-9.81) * 5.7142857²

v0= -7 m/s  ⇒ it's negative because it goes down.

v= v0 +at

v= -7 + (-9.81) * 5.7142857

v= -62.99 m/s

c) √(70² + 62.99²) = 94.17 m/s

8 0
2 years ago
A car with speed v and an identical car with speed 2v both travel the same circular section of an unbanked road. If the friction
yawa3891 [41]

Answer:

F'=\dfrac{F}{4}

Explanation:

Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.

For faster car on the road,

F=\dfrac{mv^2}{r}

v = 2v

F=\dfrac{m(2v)^2}{r}

F=4\dfrac{m(v)^2}{r}..........(1)

For the slower car on the road,

F'=\dfrac{mv^2}{r}............(2)

Equation (1) becomes,

F=4F'

F'=\dfrac{F}{4}

So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.

8 0
2 years ago
Calculate the volume of 19 kilograms of petrol if the density of petrol is 800 kg/m?​
earnstyle [38]

Answer:

i hope this will help you :)

Explanation:

mass=19kg

density=800kg/m³

volume=?

as we know that

density=mass/volume

density×volume=mass

volume=mass/density

putting the values

volume=19kg/800kg/m³

so volume=0.02375≈0.02m³

6 0
2 years ago
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