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Mrrafil [7]
3 years ago
5

If the Sun, Earth, and Moon are lined up as shown above, then the Earth would have...

Physics
2 answers:
Klio2033 [76]3 years ago
6 0

Answer: Left cheek from the sun

Explanation:

lbvjy [14]3 years ago
4 0
This is 6th grade earth science right?
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When I throw a ball in the air at what point will it have the least kinetic energy
pickupchik [31]
When you throw the ball in the air it is considered kinetic energy. Once the ball reaches its max height, it stops moving and all kinetic energy turns into potential energy. So when the ball is at its highest point.
7 0
4 years ago
Read 2 more answers
A small balloon is released at a point 150 feet away from an observer, who is on level ground. If the balloon goes straight up a
Elza [17]

Answer:

\dfrac{dz}{dt}=0.65\ ft/s

Explanation:

Given that

x= 150 ft

\dfrac{dy}{dt}= 7\ ft/s

y= 14 ft

From the diagram

z^2=x^2+y^2

When ,x= 150 ft and y= 14 ft

z^2=150^2+14^2

z=\sqrt{150^2+15^2}

z=150.74 ft

z^2=x^2+y^2

By differentiating with respect to time t

2z\dfrac{dz}{dt}= 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}

z\dfrac{dz}{dt}= x\dfrac{dx}{dt}+y\dfrac{dy}{dt}

Here x is constant that is why

\dfrac{dx}{dt}=0

z\dfrac{dz}{dt}= y\dfrac{dy}{dt}

Now by putting the values in the above equation we get

150.74\times \dfrac{dz}{dt}=14\times 7

\dfrac{dz}{dt}=\dfrac{14\times 7}{150.74}\ ft/s

\dfrac{dz}{dt}=0.65\ ft/s

Therefore the distance between balloon and observer increasing with 0.65 ft/s.

5 0
3 years ago
State the name in words and the symbol for the following:
BartSMP [9]

Answer:

<h2>Kilometer (km) and micrometer (um) respectively</h2>

Explanation:

<h3>One thousand meters is equal to one kilometer represented as km. </h3>

and

<h3>One thousandth of a meter mean 1/1000 m which implies one thousands part of a meter which is equal to micro meter and represented as um.</h3>
3 0
3 years ago
What could serve as ballast for a balloon? Help pls...thx
I am Lyosha [343]
With the advent of the plastic balloon and the beginning of the unmanned ... That would lead to a more sophisticated ballast system that uses fine steel or iron
6 0
3 years ago
The area of the piston to the master cylinder in a hydraulic braking system of a car is 0.4 square inches. If a force of 6.4 lb
Anit [1.1K]

Answer:

The force applied on one wheel during braking = 6.8 lb

Explanation:

Area of the piston (A) = 0.4 in^{2}

Force applied on the piston(F) = 6.4 lb

Pressure on the piston (P) = \frac{F}{A}

⇒ P = \frac{6.4}{0.4}

⇒ P = 16 \frac{lb}{in^{2} }

This is the pressure inside the cylinder.

Let force applied on the brake pad = F_{1}

Area of the brake pad (A_{1})= 1.7 in^{2}

Thus the pressure on the brake pad (P_{1}) =  \frac{F_{1} }{A_{1} }

When brake is applied on the vehicle the pressure on the piston is equal to pressure on the brake pad.

⇒ P = P_{1}

⇒ 16 = \frac{F_{1} }{A_{1} }

⇒ F_{1} = 16 × A_{1}

Put the value of A_{1} we get

⇒ F_{1} = 16 × 1.7

⇒ F_{1} = 27.2 lb

This the total force applied during braking.

The force applied on one wheel = \frac{F_{1} }{4} = \frac{27.2}{4} = 6.8 lb

⇒ The force applied on one wheel during braking.

7 0
3 years ago
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