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Elis [28]
3 years ago
8

I will give 50 pts and brainliest

Physics
1 answer:
ipn [44]3 years ago
8 0

Answer:

Limiting the use of fossil fuels such as oil, carbon and natural gas and replacing them with renewable and cleaner sources of energy, all while increasing energy efficiency

Explanation:

You might be interested in
What are two ways an engineer can build a car in order for it to accelerate faster
Ket [755]

Explanation:

Take F=ma

a = F/m

For a higher, F higher or m lower

Means higher horse power for engine or lower mass for the car

4 0
2 years ago
The force on the spring is F0 and it stores elastic potential energy PEs0. If the spring displacement is tripled to 3x0, determi
Dimas [21]

Answer:

Explanation:

Let initial extension in the spring= x₀

Force on the spring = F₀

Let spring constant = k

Fo = k x₀

Fn = 3k x₀

Fn /Fo = 3

PEs0 ( ORIGINAL)  =1/2 k x₀²

PEsn ( NEW)  =1/2 k (3x₀)²

PEsn / PEs0 = 9

7 0
3 years ago
A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release
Svetradugi [14.3K]

Answer with Explanation:

We are given that

Mass of spring,m=3 kg

Distance moved by object,d=0.6 m

Spring constant,k=210N/m

Height,h=1.5 m

a.Work done  to compress the spring initially=\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J

b.

By conservation law of energy

Initial energy of spring=Kinetic energy  of object

37.8=\frac{1}{2}(3)v^2

v^2=\frac{37.8\times 2}{3}

v=\sqrt{\frac{37.8\times 2}{3}}

v=5.02 m/s

c.Work done by friction on the incline,w_{friction}=P.E-spring \;energy

W_{friction}=3\times 9.8\times 1.5-37.8=6.3 J

8 0
3 years ago
Find the momentum of a particl with a mass of one gram moving with half the speed of light.
joja [24]

Answer:

129900

Explanation:

Given that

Mass of the particle, m = 1 g = 1*10^-3 kg

Speed of the particle, u = ½c

Speed of light, c = 3*10^8

To solve this, we will use the formula

p = ymu, where

y = √[1 - (u²/c²)]

Let's solve for y, first. We have

y = √[1 - (1.5*10^8²/3*10^8²)]

y = √(1 - ½²)

y = √(1 - ¼)

y = √0.75

y = 0.8660, using our newly gotten y, we use it to solve the final equation

p = ymu

p = 0.866 * 1*10^-3 * 1.5*10^8

p = 129900 kgm/s

thus, we have found that the momentum of the particle is 129900 kgm/s

6 0
3 years ago
A bullet B of mass mB traveling with a speed v0 = 1400 m/s ricochets off a fixed steel plate A of mass mA. Let mA ≫ mB so that i
Greeley [361]

Answer:

Rebounce angle is 345°

Rebounce speed is 989.95m/s

Explanation:

Calculate the x  component of the velocity of the bullet before impact by using the following relation:

Vbx= Vb Cos thetha

Here,  is the initial velocity of the bullet, Vo = 1400m/s and is the incidence angle of the bullet.= theta = 15°

Substituting

Vbx = Cos15 ×1400 = 1352.30m/s

Calculate the y component using the relation:

Vby = Vo Sin theta

Vby = sin 15° × 1400

Vby = 362.35m/s

The rebounce angle = 360 - incidence angle

Rebounce angle =( 360 - 15)° = 345°

The rebound speed V' = Vby - Vbx

V' = (1352.30 - 362.35)m/s

V' = 989.95 m/s

5 0
4 years ago
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