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Pani-rosa [81]
4 years ago
9

What is the acceleration of an object that has a velocity of 18 m/s and is moving in a circle of radius 30m?

Physics
2 answers:
blsea [12.9K]4 years ago
5 0
Acceleration = V^2/r = 18^2/ 30 = 10.8 m/s^2
insens350 [35]4 years ago
4 0

Well, first of all, 18 m/s is the object's speed, but not its velocity.
Velocity consists of the object's speed AND direction, and this
particular object's direction is constantly changing, since it's
moving around a circle.  So its velocity is constantly changing too.

Centripetal acceleration  =  (speed)² / (radius)

                                           =  (18 m/s)² / (30 m)

                                           = (324 m²/s²) / (30 m)

                                           =       10.8 m/s²

                                       (about  1.1 G)   
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A man pushes an 250-N crate at constant speed a distance of 30.0 m upward along a rough slope that makes an angle of 60° with th
mel-nik [20]

Answer:

6495.19 Joule

Explanation:

F = Weight of the crate = 250 N

d = Distance the cart is pushed = 30 m

θ = Angle of inclination = 60°

The weight of the crate will be resloved into two components

Fdsinθ and Fdcosθ

Work done by the force of gravity is

W = Fdsinθ

⇒W = 250×30×sin60

⇒W = 6495.19 Joule

∴ The work done by the force of gravity is 6495.19 Joule

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3 years ago
Define any two characteristics of sound​
Oliga [24]

Answer:

five characteristics: Wavelength, Amplitude, Time-Period, Frequency and Velocity or Speed

5 0
3 years ago
A closed system:
adoni [48]
The answer is b i just did the test
3 0
2 years ago
A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o
iVinArrow [24]
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: v=15 m/s
- radius of the hill: r=100 m

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car W=mg (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, m \frac{v^2}{r}, so we can write:
mg-N=m \frac{v^2}{r} (1)
By rearranging the equation and substituting the numbers, we find N:
N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
mg=m \frac{v^2}{r}
from which we find
v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s
7 0
4 years ago
7. A beam of light converges at a point P. Now a lens is placed in the path of the convergent
Rudiy27

Answer:

Lens at a distance = 7.5 cm

Lens at a distance = 6.86 cm  (Approx)

Explanation:

Given:

Object distance u = 12 cm

a) Focal length = 20 cm

b) Focal length = 16 cm

Computation:

a. 1/v = 1/u + 1/f

1/v = 1/20 + 1/12

v = 7.5 cm

Lens at a distance = 7.5 cm

b.  1/v = 1/u + 1/f

1/v = 1/16 + 1/12

v = 6.86 cm  (Approx)

Lens at a distance = 6.86 cm  (Approx)

5 0
3 years ago
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