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-BARSIC- [3]
2 years ago
6

>

Physics
1 answer:
tester [92]2 years ago
4 0

The pressure exerted on the block on the ground in N/m² is 200N/m².

<h3>What is pressure?</h3>

The pressure is the amount of force applied per unit area.

Given is a 5000 Newton block rests on the ground over 25 m² of area.

Pressure p = Force/Area

Put the values, we get

p = 5000 /25

p = 200 N/m²

Hence, pressure exerted on the block on the ground is 200 N/m².

Learn more about pressure.

brainly.com/question/12971272

#SPJ1

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How do we use energy transformation in our daily lives?
olga55 [171]

Answer:hat are some examples of energy transformation?

The Sun transforms nuclear energy into heat and light energy.

Our bodies convert chemical energy in our food into mechanical energy for us to move.

An electric fan transforms electrical energy into kinetic energy.

Explanation:

4 0
2 years ago
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A building is 512 m high. What must be the minimum water pressure in a pipe at ground level in order to get water out of a tap i
vivado [14]

Answer:

Pressure =  5 x 10⁶ Pa

Explanation:

Given:

Height of building = 512 m

Find:

Pressure

Computation:

P2 = P1+dgh

P2 = 1 + (1000)(9.8)(512)

P2 = 51.2 atm

Pressure =  5 x 10⁶ Pa

6 0
3 years ago
In the inclined plane the objects that are thrown from a more inclined place go faster?
Mamont248 [21]
Assuming the objects roll down the inclined plane, yes.
If the object never touches the plane, then no.
7 0
3 years ago
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A particle with a charge of +4.20nC is in a uniform electric field E⃗ directed to the left. It is released from rest and moves t
Morgarella [4.7K]

Answer:

(A). The work done is 1.50\times10^{-6}\ J.

(B). The potential of the starting point with respect to the endpoint is 357.14 V.

(C). The magnitude of E is 5952.38 N/C.

Explanation:

Given that,

Charge = 4.20 nC

Distance = 6.00 cm

Kinetic energy K.E=1.50\times10^{-6}\ J

The particle start from rest.

So, the initial kinetic energy i zero.

(A). We need to calculate the work by the electric force

Using formula of work done

W = \Delta K.E

W=K.E_{f}-K.E_{i}

Put the value into the formula

W= 1.50\times10^{-6}-0

W=1.50\times10^{-6}\ J

The work done is 1.50\times10^{-6}\ J.

(B). We need to calculate the potential of the starting point with respect to the endpoint

We know that.

Change in potential energy = change in kinetic energy

\Delta P.E=\Delta K.E

So, U = 1.50\times10^{-6}

Using formula of potential

V=\dfrac{U}{q}

Put the value into the formula

V=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}}

V=357.14\ V

The potential of the starting point with respect to the endpoint is 357.14 V.

(C). We need to calculate the magnitude of E

Using formula of work done

W=F\times r....(I)

Using formula of force

F=qE

Put the value in the equation (I)

W=qE\times r

E=\dfrac{W}{q\times r}

Put the value into the formula

E=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}\times6.00\times10^{-2}}

E=5952.38\ N/C

The magnitude of E is 5952.38 N/C.

Hence, This is the required solution.

7 0
3 years ago
30
BARSIC [14]

Answer:

The correct answer is Tor buni aga daka pagli Tor buda mu tor buni kaitam chai ami

5 0
2 years ago
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