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rjkz [21]
3 years ago
9

On the asteroid Cere, the acceleration due to gravity is about 0.27 m/s^2. How long would it take a ball to fall to the ground i

f dropped from a height of 90 m?
Answer Choices: A. 30.77s B. 28.4s C. 27,12s D. 29.79s E. 31.84s F. 25.82s
Physics
1 answer:
KiRa [710]3 years ago
7 0

Answer:

F. 25.82 s

Explanation:

Given:

Δy = 90 m

v₀ = 0 m/s

a = 0.27 m/s²

Find: t

Δy = v₀ t + ½ at²

90 m = (0 m/s) t + ½ (0.27 m/s²) t²

t = 25.82 s

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A stationary speed gun emits a microwave beam at 2.10*10^10Hz. It reflects off a car and returns 1030 Hz higher. What is the spe
KIM [24]

Answer: V = 15 m/s

Explanation:

As  stationary speed gun emits a microwave beam at 2.10*10^10Hz. It reflects off a car and returns 1030 Hz higher. The observed frequency the car will be experiencing will be addition of the two frequency. That is,

F = 2.1 × 10^10 + 1030 = 2.100000103×10^10Hz

Using doppler effect formula

F = C/ ( C - V) × f

Where

F = observed frequency

f = source frequency

C = speed of light = 3×10^8

V = speed of the car

Substitute all the parameters into the formula

2.100000103×10^10 = 3×10^8/(3×10^8 -V) × 2.1×10^10

2.100000103×10^10/2.1×10^10 = 3×108/(3×10^8 - V)

1.000000049 = 3×10^8/(3×10^8 - V)

Cross multiply

300000014.7 - 1.000000049V = 3×10^8

Collect the like terms

1.000000049V = 14.71429

Make V the subject of formula

V = 14.71429/1.000000049

V = 14.7 m/s

The speed of the car is 15 m/s approximately.

8 0
2 years ago
an empty density bottle weighed 23.5 and 48.4g when filled with water .The empty bottle was partially filled with sand until and
gavmur [86]

Answer:

Density of Sand is 2.653g/cm^{3}.

Explanation:

Given Empty Density bottle weighs 23.5gm(W=23.5gm)

Weight of bottle when completely filled water=48.4gm

So amount of water required to fill the bottle=Weight of bottle filled with water-W

Amount of water required to fill the bottle(w_{max})=48.4gm-23.5gm

w_{max}=24.9g

Since we know density of water d_{w} =1g/cm^{3} and w_{max}

We can calculate volume of empty space in the bottle(V).

w_{max}=d_{w}\timesV

V=\frac{w_{max} }{d_{w} }

V=\frac{24.9}{1}

V=24.9 g/cm^{3}

Now bottle is partially filled with sand,and weight of bottle is (w_{s})36.5gm

So,

Amount of sand added (m_{s})=36.5-Weight of the bottle

m_{s}=13g

After filling the bottle with water again,the weight of the bottle becomes (W_{2}=56.5g)

Therefore,

amount of water added to the bottle of sand in grams = W_{2}-36.5gm

amount of water added =56.5g-36.5g

amount of water added =20g

As the density of water = 1g/cm^{3}

Amount of water (in grams )=Volume of water occupied

20=volume of water added

Therfore volume of water added to the sand filled bottle(V_{w})=20cm^{3}

As we know the total volume of the water bottle(V),

Volume of the sand occupied in the water bottle=V-V_{w}

V_{s}=24.9g-20g

V_{s}=4.9g

We know,

Density=Mass/Volume

Therefore,

density of sand = \frac{m_{s} }{V_{s} }

density of sand =\frac{13}{4.9}

density of sand = 2.653g/cm^{3}

8 0
2 years ago
Why are weight and mass used synonymously on earth?
Lisa [10]
Weight and mass are used synonymously on earth because the value of g is constant on the earth because the weight of a body is the amount of force exerted by the Earths gravity on an object of finite mass. And, the mass of an object gives the amount of matter in the body and is measured in kilograms. 
6 0
3 years ago
The liver’s role is to:
Lelu [443]
D. The liver's role is to remove harmful substances from the blood.
7 0
3 years ago
n a car lift used in a service station, compressed air exerts a force on a small piston that has a circular cross section of rad
Oksi-84 [34.3K]

Answer: 1477.78 N

Explanation:

Let's assume that the cross sectional area of the smaller piston be A1

let's also assume the cross sectional area of the larger piston be A2

We assume the force applied to the smaller piston be F1

We also assume the force applied to the larger piston be F2

we then use the formula

F1/A1 = F2/A2

From our question,

The radius of the smaller piston is 5 cm = 0.05 m

The radius of the larger piston is 15 cm = 0.15 m

The force of the larger piston is 13300 N

The force of the smaller piston is unknown = F

A1 = πr² = 3.142 * 0.05² = 0.007855 m²

A2 = πr² = 3.142 * 0.15² = 0.070695 m²

F1/0.007855 = 13300/0.070695

F1 = (13300 * 0.007855) / 0.070695

F1 = 104.4715 / 0.070695

F1 = 1477.78 N

Thus, the force the compressed air must exert is 1477.78 N

5 0
3 years ago
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