4. The result is that a(n)<br> electric current is induced<br> PLA

Tremendous energy is released by nuclear reactions is a measure of the energy required to bond the nucleus together. Where does

Harlamova29_29 [7]

Answer:

Fission is the opposite of fusion and releases energy only when heavy nuclei are split. As noted in Fusion, energy is released if the products of a nuclear reaction have a greater binding energy per nucleon than the parent nuclei.

The amount of energy released during nuclear fission is millions of times more efficient per mass than that of coal considering only 0.1 percent of the original nuclei is converted to energy. Daughter nucleus, energy, and particles such as neutrons are released as a result of the reaction

5 0

2 years ago

What provides the centripetal force needed to keep Earth in orbit?

kkurt [141]

The centripetal force needed to keep earth in orbit is gravity.

3 0

2 years ago

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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

2 years ago

How much heat is needed to raise the temperature of a 100.0g of water by 85.0 c?

KatRina [158]

Answer:

35.7kJ

Explanation:

we can calculate the amount of heat energy required , using this formula

Q = mcθ

where.

Q = heat energy (Joules, J)

m = mass of a substance (kg)

c = specific heat capacity (units $Jkg^{-1} C^{-1}$ )

θ = change in temperature (Celcius,C or Kelvin K)

Assume Specific heat capacity (c) of water = $4200Jkg^{-1} C^{-1}$

mass =0.1 kg

$Q=0.1 kg*4200Jkg^{-1} C^{-1}*85C\\=35700J\\=35.7kJ$

3 0

2 years ago

A 2 kg book is pushed from rest to a final velocity of 3 m/s. The book travels 2 m. How much force was the book pushed with

Studentka2010 [4]

Explanation:

first you have to find accelerarion, it is given that the initial velocity(u) is 3 m/s, distance travelled(s) be 2m finall it came to rest so final velocity be 0m/s

now using the 3rd law of motion

v^2=u^2+2as

0=9+2a2

a= -9/4m/s^2

now force=mass×accelration

=2kg×(-9/4)m/s^2

=4.5 N

4.5 newton force applied on the book!

✌️:)

6 0

2 years ago

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4. The result is that a(n) electric current is induced PLA