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3 years ago

A 20 kg truck drives in a circle of radius 4 m at 10m/s. What is the centripetal acceleration of the truck?

Physics

1 answer:

asambeis [7]3 years ago

4 0

Answer:

B. 25 m/s/s

Explanation:

Centripetal acceleration is the square of the tangential velocity divided by the radius of curvature.

a = v² / r

Given v = 10 m/s and r = 4 m:

a = (10 m/s)² / 4 m

a = 25 m/s²

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a train is moving at a constant speed on a surface and inclined at 10 degrees with the horizontal travels a distance of 400 m in

BabaBlast [244]

Answer:

13.9 m/s

Explanation:

The velocity of the train along the inclined surface is given by the ratio between the distance travelled, d, and the time taken, t:

$v=\frac{d}{t}$

where we have

d = 400 m

t = 5 s

Substituting,

$v=\frac{400 m}{5 s}=80 m/s$

This is the velocity of the train along the inclined plane. Now we can find the vertical component of the velocity by using the formula:

$v_y = v sin \theta$

where $\theta=10^{\circ}$ is the angle of the slope. Substituting, we find

$v_y = (80 m/s)(sin 10^{\circ})=13.9 m/s$

3 0

3 years ago

A model airplane moves twice as fast as another identical model airplane. compared with the kinetic energy of the slower airplan

Papessa [141]

Twice as fast, so multiply the kinetic energy of the slower plane by two.

6 0

3 years ago

What is the net force of the object below

tino4ka555 [31]

The answer to this question is C

7 0

3 years ago

Read 2 more answers

Find the electric force acting on an alpha particle in a horizontal electric field of 600N/C

Sergeeva-Olga [200]

Since an alpha particle has 2 protons and no negative particles (electrons) to balance the net charge, its charge is
Q=2(1.6e-19)=3.2e-19C.
The force on a charged particle is F=QE so
(3.2e-19C)(600N/C)=1.92e-16N

6 0

3 years ago

A bird watcher meanders through the woods, walking 1.62 km due east, 0.246 km due south, and 3.08 km in a direction 50.4 ° north

Liula [17]

Answer:

Displacement CO=2.15 Km

Average velocity = 1.99 Km/h

Explanation:

Given

Total time = 1.08 h

OA= 1.62 Km

AB=0.246 Km

BC=3.08 Km

θ=50.4°

CE=3.08 sin50.4° =2.37 Km

BE=3.08 cos50.4° =1.96 Km

CE=AB+OD

OD=CE-AB

OD=2.37-0.246

OD=2.124 Km

BE=CD+OA

CD=BE-OA

CD=1.96-1.62

CD=0.34 Km

Now CO

$CO=\sqrt{CD^2+OD^2}$

$CO=\sqrt{0.34^2+2.124^2}$

CO=2.15 Km

So the displacement CO=2.15 Km

Average velocity =Displacement/Time

Average velocity = 2.15/1.08 Km/h

Average velocity = 1.99 Km/h

8 0

4 years ago