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vampirchik [111]
3 years ago
5

An ostrich can run at speeds of up to 72 kilometers per hour. How long will it take an ostrich to run 1.5 kilometers at this top

speed?
Physics
1 answer:
tatiyna3 years ago
6 0
S=d/t
T=d/s
= 1.5/72
= 0.02 hours
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A pen contains a spring with a spring constant of 257 N/m. When the tip of the pen is in its retracted position, the spring is c
Mandarinka [93]

Explanation:

Below is an attachment containing the solution.

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3 years ago
A wave emitted from a source has a frequency of 10 Hz and wavelength 2.5 m. How much time will it take to reach a person located
const2013 [10]

Answer:

time taken by the wave to reach the person is 0.2 s

Explanation:

As we know that the speed of the wave is given as

v = \lambda f

here we know that the wavelength of the wave is

\lambda = 2.5 m

f = 10 Hz

now speed of the wave is given as

v = 10(2.5)

v = 25 m/s

Now time taken by the wave to reach 5 m distance is

t = \frac{L}{v}

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4 0
3 years ago
Which statement most completely describes a Lewis electron dot diagram? A.
andrezito [222]

Answer:

i do not know

Explanation:

5 0
2 years ago
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he membrane that surrounds a certain type of living cell has a surface area of 6.0 x 10-9 m2 and a thickness of 1.6 x 10-8 m. As
k0ka [10]

Answer:

1.54481175\times 10^{-12}\ C

Explanation:

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

A = Area = 6\times 10^{-9}\ m^2

d = Thickness = 1.6\times 10^{-8}\ m

k = Dielectric constant = 5.4

V = Voltage = 86.2 mV

Charge is given by

Q=CV\\\Rightarrow Q=k\epsilon\dfrac{A}{d}V\\\Rightarrow Q=5.4\times 8.85\times 10^{-12}\times \dfrac{6\times 10^{-9}}{1.6\times 10^{-8}}\times 86.2\times 10^{-3}\\\Rightarrow Q=1.54481175\times 10^{-12}\ C

The charge on the outer surface is 1.54481175\times 10^{-12}\ C

4 0
3 years ago
Calculate the translational speed of a cylinder when it reaches the foot of an incline 7.05 mm high. Assume it starts from rest
mestny [16]

Height is 7.05 m and not 7.05 mm

Answer:

9.603 m/s

Explanation:

We are dealing with rotation, so velocity of centre of mass is given by;

v_cm = Rω

Since we are working with a solid cylinder, moment of inertia of the cylinder is; I = ½mR²

Since it is rolled from the top to the bottom, at the top it will have potential energy(mgh) while at the bottom it will have kinetic energy (rotational plus translational kinetic energy).

Using conservation of energy, we have:

P.E = K.E_t + K.E_r

Formula for rotational and kinetic energy here are;

K.E_t = ½mv²

K.E_r = ½Iω²

mgh = ½mv² + ½Iω²

Since we want to find translational speed(v), let's get rid of ω.

Earlier, we saw that v_cm = Rω

Thus; ω = v/R

Also, we know that I = ½mR².

Thus;

mgh = ½mv² + ½(½mR²)(v/R)²

This gives;

mgh = ½mv² + ¼mv²

Divide through by m to get;

gh = v²(½ + ¼)

gh = ¾v²

Making v the subject gives;

v = √(4gh/3)

v = √((4 × 9.81 × 7.05)/3)

v = 9.603 m/s

6 0
3 years ago
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