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Dvinal [7]
2 years ago
8

Data from a survey:

Physics
1 answer:
gayaneshka [121]2 years ago
4 0
C I think the answer is true
You might be interested in
Starting from rest, a car takes 2.4s to travel 15m. Assuming a constant acceleration, how long will it take the car to travel th
Alborosie

Answer:

T = 2.4 + 2.4 = 4.8 [s]

Explanation:

In order to solve this problem, we must use the following kinematics equation and calculate the acceleration value.

x=x_{o} +v_{o}*t+(\frac{1}{2})*a*t^{2}

Vo = inital velocity = 0

x - xo = 15 [m]

t = time = 2.4 [s]

15 = 0.5*a*(2.4)^2

a = 5.208 [m/s^2]

We can use the same equation to find the time.

30 = 15 + 0.5*(5.208)*t^2

t = 2.4 [s]

T = 2.4 + 2.4 = 4.8 [s]

7 0
3 years ago
8.
Arisa [49]

Answer:

can you translate that plz

Explanation:

5 0
2 years ago
Read 2 more answers
What is a centripetal acceleration of a point on a bicycle wheel of a radius of 0.70 m when a bike is moving 8.0 m/s
Furkat [3]

Answer:

The acceleration of a point on the wheel is 11.43 m/s² acting radially inward.

Explanation:

The centripetal acceleration acts on a body when it is performing a circular motion.

Here, a point on the bicycle is performing circular motion as the rotation of the wheel produces a circular motion.

The centripetal acceleration of a point moving with a velocity v and at a distance of r from the axis of rotation is given as:

a=\frac{v^2}{r}

Here, v=8\ m/s,r=0.70\ m

∴ a=\frac{8}{0.70}=11.43\ m/s^2

Therefore, the acceleration of a point on the wheel is 11.43 m/s² acting radially inward.

3 0
3 years ago
A 0.20 kg mass is oscillating at a small angle from a light string with a period of 0.78 s.
Scorpion4ik [409]

Answer:

L = 15 cm

Explanation:

T = 2π√(L/g)

L = g(T/2π)²

L = 9.8(0.78/2π)²

L = 0.151027... m

L = 15 cm

7 0
2 years ago
Read 2 more answers
A glass tube 1 mm in diameter is dipped into glycerin. The density of the glycerin is 1260 kg/m3, surface tension is 6.3x10-2 N/
Sunny_sXe [5.5K]

Answer:

The capillary rise of the glycerin is most nearly  y  =  0.0204 \ m

Explanation:

From the question we are told that

  The diameter of the glass tube is  d =  1 \ mm =  0.001 \ m

   The density of glycerin is  \rho =  1260 \ kg /m^3

   The surface tension of the glycerin is \sigma   =  6.3 *10^{-2} \ N /m

The capillary rise of the glycerin is mathematically represented as

       y  =  \frac{4 * \sigma  *  cos (\theta )}{ \rho * g *  d}

substituting value  

       y  =  \frac{4 * 6.3 *10^{-2}  *  cos (0 )}{ 1260 * 9.8 *  0.001}

      y  =  0.0204 \ m

Therefore the height  of the glass tube  the glycerin was able to cover is

y  =  0.0204 \ m  

4 0
3 years ago
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