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Xelga [282]
3 years ago
7

A car travels from point A to Be in 3 hours and returns back to point A in 5 hours .Points A and B are 150 miles apart along str

aight highway .Find the average velocity for A to B ,B to A and for the entire trip in miles per hour
Physics
1 answer:
Rainbow [258]3 years ago
3 0

Explanation:

Given that,

The time taken from A to B is 3 hours

The time from B to A = 5 hours

The total distance between A and B is 150 miles

The average velocity from A to B,

v=\dfrac{150}{3}=50\ mph

The average velocity from B to A,

v=\dfrac{150}{5}=30\ mph

As the velocity of an object is a vector quantity. It means when the car reaches the initial point, the displacement is 0. So, the average velocity is equal to 0.

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An apple falls from an apple tree growing on a 20° slope. The apple hits the ground with an impact velocity of 16.2 m/s straight
EleoNora [17]

Apple hits the surface with speed 16.2 m/s

The angle made by the apple velocity with normal to the incline surface is given as 20 degree

now the component of velocity which is parallel to the surface and perpendicular to the surface is given as

v_{perpendicular} = v cos20

v_{parallel} = v sin20

so here we have

v_{parallel} = 16.2 sin20

v_{parallel} = 5.5 m/s

<em>so its velocity along the incline plane will be 5.5 m/s</em>

7 0
4 years ago
How much power should a braked 1.5t car have to be braked to reduce its speed from 30m / s to 10m / s in 5s?​
grandymaker [24]

Answer:

-120000 W

Explanation:

Power = change in energy / time

P = ΔE / t

P = (½ mv₂² − ½ mv₁²) / t

P = m (v₂² − v₁²) / (2t)

Given m = 1.5 t = 1500 kg, v₂ = 10 m/s, v₁ = 30 m/s, and t = 5 s:

P = (1500 kg) ((10 m/s)² − (30 m/s)²) / (2 × 5 s)

P = -120000 W

7 0
4 years ago
The crankshaft in a race car goes from rest to 3000 rpm in 3.0 s . (a) What is the crankshaft's angular acceleration?
Talja [164]

Answer:

75 rotations

Explanation:

f0 = 0, f = 3000 rpm = 50 rps, t = 3 s

(a) use first equation of motion for rotational motion

w = w0 + α t

2 x 3.14 x 50 = 0 + α x 3

α = 104.67 rad/s^2

(b) Let θ be the angular displacement

use second equation of motion for rotational motion

θ = w0 t + 1/2 α t^2

θ = 0 + 0.5 x 104.67 x 3 x 3

θ = 471.015 rad

The angle turn in one rotation is 2 π radian.

Number of rotation = 471.015 / (2 x 3.14) = 75 rotations

7 0
3 years ago
A man pushes his child in a grocery cart. The total mass of the cart and child is 30.0 kg. If the force of friction on the cart
Ber [7]
Newton's second law states that the resultant of the forces applied to an object is equal to the product between the object's mass and its acceleration:
\sum F = ma
where in our problem, m is the mass the (child+cart) and a is the acceleration of the system.

We are only concerned about what it happens on the horizontal axis, so there are two forces acting on the cart+child system: the force F of the man pushing it, and the frictional force F_f acting in the opposite direction. So Newton's second law can be rewritten as
F-F_a = ma
or
F=ma + F_f

since the frictional force is 15 N and we want to achieve an acceleration of a=1.50 m/s^2, we can substitute these values to find what is the force the man needs:
F=(30 kg)(1.5 m/s^2)+15 N=60 N
8 0
4 years ago
An object moves in a circle at a constant speed of 1.0 m/s. The radius of the circle is 1.0 m. If a force of 1.0 N acts toward t
Vlad1618 [11]

Answer:5

Explanation:

Given

speed of object v=1\ m/s

radius of circle r=1\ m

Force towards the center F=1\ N

Work done is given by the dot product of Force and displacement

and we know know displacement of the object is along the circle which is perpendicular to the force acting therefore Work done will be zero

W=F\cdot s\cos 90

W=0

4 0
4 years ago
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