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coldgirl [10]
3 years ago
12

A person finds it easier to lift an anchor through the water than lift it into the boat through the air. Why?

Physics
1 answer:
Tamiku [17]3 years ago
3 0

Answer:

because underwater a force called buoyant force of upthrust applies in the objects in water in the upward direction

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While John is traveling along an interstate highway, he notices a 150 mi marker as he passes through town. Later John passes a 1
Lerok [7]

Answer:

Part a)

distance = 112 miles

Part b)

current position = 112 miles from the position of town

Explanation:

Part a)

Since the distance marker is showing the distance between the town and the position of john at all time

so here we have

d = 112 miles

Part b)

Current position of John is given as

r = 112 miles

from the position of the town

7 0
4 years ago
Read 2 more answers
Generation of electricity in coal-burning power plants and nuclear power plants both involve _______.
kolbaska11 [484]

Answer:

Heating water to produce steam which drives a turbine

Explanation:

Generation of electricity in coal-burning power plants and nuclear power plants both involve heating water to produce steam which drives a turbine.

5 0
3 years ago
How might the changes in tide affect a fishing community?
artcher [175]

Answer:

THEY WOULD FIND DED FISH:)))))

Explanation:

6 0
3 years ago
Compare and contrast series and parallel circuits?
dalvyx [7]

In a series circuit, a common current flows through all the components of the circuit. While in a parallel circuit, a different amount of current flows through each parallel branch of the circuit. Whereas in the parallel circuit, the same voltage exists across the multiple components in the circuit.

Hope It Helps!

6 0
3 years ago
A flat sheet is in the shape of a rectangle with sides of lengths 0.400 m and 0.600 m. The sheet is immersed in a uniform electr
Katen [24]

Answer:

ФE = 9.403W

Explanation:

In order to calculate the magnitude of the electric flux trough the sheet, you use the following formula:

\Phi_E=\vec{A}\cdot \vec{E}=AEcos\theta       (1)

A: area of the rectangular sheet = (0.400m)(0.600m) = 0.24m^2

E: magnitude of the electric field = 95.0N/C

θ: angle between the direction of the electric field and the normal to the surface of the sheet

You replace the values of the parameters in the equation (1):

\Phi_E=(0.24m^2)(95.0N/m)cos(20\°)=9.304W

The magnitude of the electric flux is trough the sheet is 9.403W

5 0
3 years ago
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