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Lena [83]
3 years ago
10

5. Calculate how long in seconds it will take for a light pulse to travel from earth to the moon. The distance from earth to moo

n is 3.84 x 105 km. Remember that light travels at 186,000 mi/s. (4 pts)
Physics
1 answer:
solniwko [45]3 years ago
3 0

Answer:

0.13 seconds

Explanation:

Since 1 Km = 0.621 miles

3.84 x 105 km = 3.84 x 105 × 0.621 = 23846.4 miles

Speed = distance/time

time= distance/speed

Time= 23846.4/186,000

Time= 0.13 seconds

You might be interested in
A disadvantage of some geophysical techniques based on magnetism is that their use in urban areas may be subject to distortion r
julsineya [31]

Answer:

Resistivity Method

Explanation:

A disadvantage of some geophysical techniques based on magnetism is that their use in urban areas may be subject to distortion resulting from power lines and metals in the immediate vicinity. One method that can be used in such areas relies on soil moisture and the ground's resistance to electricity and is known as Resistivity.

It is usually carried out by injecting a small electrical current through the earth and measured a subtle  sub-surface change in resistance over a given area.

3 0
3 years ago
At a processing facility outside of Detroit, a 7.0 kg box of goat cheese, initially at rest, is given a push up a smooth, inclin
Firlakuza [10]

Answer:

d = 5.9 m

Explanation:

During the initial push, there are two forces acting on the box along  the ramp: the component of gravity force along the ramp (directed to the bottom of the ramp), and the pushing force (up the ramp).

As we have as a given, the time during which the pushing force is applied, we can use Newton's 2nd Law, expressed in its original form, as follows:

Fnet *Δt = Δp = m* (vf-v₀) (1)

Fnet, in this case, is the difference between Fapp, and the projection of Fg along the ramp, which is equal to Fg times the sinus of  the angle of the ramp with respect to the horizontal.

We choose as positive, the direction up the ramp, so we can write the follwing equation:

⇒ Fnet = Fapp - Fg*sin θ = 140 N - (7kg*9.8m/s²*sin 30º) = 140 N - 34.3 N

⇒ Fnet = 105.7 N

Replacing in (1):

105.7 N * 0.5 s = 7 kg* vf (v₀=0, as the box starts from rest)

Solving for vf:

vf = 105.7 N* 0.5 s / 7 kg = 7.6 m/s

When the push ends, the only force remaining along  the ramp, is the component of Fg that we have already obtained, that will cause the box to have a deceleration, which we can find out aplying Newton's 2nd Law, as follows:

m*g*sin 30º = m*a

As we have defined as positive direction the one up the ramp, a will be negative (as it is slowing down the box) , and can be calculated as follows:

a = -34.3 N / 7 kg = -4.9 m/s²

As this value is constant, we can use any kinematic equation in order to get  the distance traveled, farther the point where it disappeared the influence of the pushing force:

vf² - v₀² = 2*a*d

As we know that finally the box will come momentarily at rest (before falling under the influence of  gravity) , we have vf =0:

⇒ -v₀² = 2*a*d

For this part, v₀, is just the value for vf, that we got above:

v₀= 7.6 m/s

⇒ -(7.6)² =2*(-4.9 m/s²)*d

Solving finally for d (the answer we are looking for):

d = (7.6)² (m/s)² / 2*4.9 m/s² = 5.9 m

8 0
3 years ago
small plastic container, called the coolant reservoir, catches the radiator fluid that overflowswhen the automobile engine becom
Ilia_Sergeevich [38]

Answer:

0.53 quart

Explanation:

The volume expansion of the coolant is gotten from ΔV = VγΔθ where ΔV   = change in volume of the coolant, V = initial volume of coolant = 15 quart, γ = coefficient of volume expansion of coolant = 410 × 10⁻⁶ /°C and Δθ = temperature change = θ₂ - θ₁ where θ₁ = initial temperature of coolant = 6 °C and θ₂ = final temperature of coolant = 92 °C. So, Δθ = θ₂ - θ₁ = 92 °C - 6 °C = 86 °C

Since, ΔV = VγΔθ

substituting the values of the variables into the equation, we have

ΔV = VγΔθ

ΔV = 15 × 410 × 10⁻⁶ /°C × 86 °C

ΔV = 528900 × 10⁻⁶ quart

ΔV = 0.528900 quart

ΔV ≅ 0.53 quart

Since the change in volume of the coolant equals the spill over volume, thus the overflow from the radiator will spill into the reservoir when the coolant reaches its operating temperature of 92 °C is 0.53 quart.

4 0
3 years ago
A rocket with a mass of 62,000 kg (including fuel) is burning fuel at the rate of 150 kg/s and the speed of the exhaust gases is
mezya [45]

Answer:

h≅ 58 m

Explanation:

GIVEN:

mass of rocket M= 62,000 kg

fuel consumption rate =  150 kg/s

velocity of exhaust gases v= 6000 m/s

Now thrust = rate of fuel consumption×velocity of exhaust gases

=6000 × 150 = 900000 N

now to need calculate time t = amount of fuel consumed÷ rate

= 744/150= 4.96 sec

applying newton's law

M×a= thrust - Mg

62000 a=900000- 62000×9.8

acceleration a= 4.71 m/s^2

its height after 744 kg of its total fuel load has been consumed

h= \frac{1}{2}at^2

h= \frac{1}{2}4.71\times4.96^2

h= 58.012 m

h≅ 58 m

4 0
3 years ago
3. If a pipe with flowing water has a cross-sectional area nine times greater at point 2 than at point 1, what would be the rela
shusha [124]

Answer:

The velocity of the flow at point 1 is nine times greater than the velocity at point 2.

Explanation:

We know that the same volume of water entering through point 1 must exit through point 2.

The volume of water per unit of time or the volumetric flow is defined by:

V_{flow}=v*A\\ where:\\V_{flow}= flow [m^3/s]\\v=velocity[m/s]\\A=area [m^2]

If  A_{2}=9*A_{1}  \\

therefore

v_{1} *A_{1} = v_{2} *A_{2} \\v_{1} *A_{1} = v_{2} *9A_{1} \\\\v_{1} = 9v_{2}  \\

4 0
4 years ago
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