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Alik [6]
3 years ago
9

If a toy car has a centripetal acceleration of 50 m/s2 and was making the turn at 10 m/s. what was his radius

Physics
1 answer:
S_A_V [24]3 years ago
3 0

Answer:

Explanation:

The centripetal acceleration is expressed as;

a = v²/r

a is the acceleration = 50m/s²

v is the velocity = 10m/s

r is the radius

To get the radius

r = v²/a

r = 10²/50

r = 100/50

r = 2m

Hence its radius is 2m

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a spring with a constant of 80N/m is stretched by a force of 240N. how much the displacement of the spring from equilibrium?
Inessa05 [86]

Answer:

1200N/m

Explanation:

given parameters:

force on the motorcycle spring is 240N

Extension 2cm or 0.02m

unknown _

spring constant:

:?

solution:

to a spring a force applied is given as :

f=ke

f is applied as force

k is spring constant

e is the Extension

240= kx0.02

k=1200N/m

8 0
3 years ago
The velocity of the transverse waves produced by an earthquake is 8.9 km/s, and that of the longitudinal waves is 5.1 km/s. A se
Brrunno [24]

Answer: The distance is 723.4km

Explanation:

The velocity of the transverse waves is 8.9km/s

The velocity of the longitudinal wave is 5.1 km/s

The transverse one reaches 68 seconds before the longitudinal.

if the distance is X, we know that:

X/(9.8km/s) = T1

X/(5.1km/s) = T2

T2 = T1 + 68s

Where T1 and T2 are the time that each wave needs to reach the sesmograph.

We replace the third equation into the second and get:

X/(9.8km/s) = T1

X/(5.1km/s) = T1 + 68s

Now, we can replace T1 from the first equation into the second one:

X/(5.1km/s) = X/(9.8km/s) + 68s

Now we can solve it for X and find the distance.

X/(5.1km/s) - X/(9.8km/s) = 68s

X(1/(5.1km/s) - 1/(9.8km/s)) = X*0.094s/km= 68s

X = 68s/0.094s/km = 723.4 km

6 0
4 years ago
In your daily life you come across a range of motion in which acceleration is in the direction of motion
irakobra [83]
That makes no sense to me somehow
3 0
3 years ago
A laser of wavelength 720 nm illuminates a double slit where the separation between the slits is 0.22 mm. Fringes are seen on a
kumpel [21]

Answer:

The appropriate solution is "2.78 mm".

Explanation:

Given:

\lambda = 720 \ nm

or,

  = 720\times 10^{-9} \ m

D=0.85 \ m

d = 0.22 \ mm

or,

  =0.22 \times 10^{-3} \ m

As we know,

Fringe width is:

⇒ \beta=\frac{\lambda D}{d}

hence,

Separation between second and third bright fringes will be:

⇒ \theta=\beta=\frac{\lambda D}{d}

       =\frac{720\times 10^{-9}\times 0.85}{0.22\times 10^{-3}}

       =2.78\times 10^{-3} \ m

or,

       =2.78 \ mm

8 0
3 years ago
Once an object enters orbit, what keeps the object moving sideways?
zubka84 [21]

Answer: Inertia!!

Explanation: I just completed the edg quiz and got that answer correct! Hope its not too late for you!

7 0
3 years ago
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