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3 years ago

The world's demand for fuel is ever increasing. Identify the fuel type predicted to run out within the next hundred years.

Physics

1 answer:

Step2247 [10]3 years ago

6 0

Answer = A) fossil fuels

B and C are renewable types of energy
D is a type of fossil fuel, just like oil

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a crane does 9,500 J of work to lift a crate straight up using a force of 125 N. how high does the crane lift the crate?

Lina20 [59]

Answer:

The crane lifts the crate up to 76 m high

Explanation:

Work Done by a Force

The work done by a force of magnitude F that displaces an object by a distance y is given by

$W=F.y$

We know a crane does a work of 9,500 Joule to lift a crate and is using a force of 125 Nw.

The above equation can be solved to know the value of y in terms of the work and the force:

$\displaystyle y=\frac{W}{F}$

Plugging in the given values

$\displaystyle y=\frac{9,500}{125}$

$y=76\ m$

The crane lifts the crate up to 76 m high

7 0

2 years ago

What is the distance from the earth's center to a point outside the earth where the gravitational acceleration due to the earth

Crazy boy [7]

R2^ 2 / R1 ^2 = g1 / g2 = 38

R2 = R1 x √38 = 6.1644* R1 

R2 = 6.1644 x 6378 000 = 39316632.5 m

8 0

3 years ago

An electric dipole of dipole moment 'p' is placed in the position of stable equilibrium in a uniform field of intensity 'E'. The

larisa86 [58]

Answer:

Torque by electric dipole = pEcos thita

5 0

3 years ago

The ice skaters partner liftes her up a distance of 1 m work done or not work done

SOVA2 [1]

Answer:

Work done.

Explanation:

The skater who lifts has to overcome the partner's weight. When lifted up by 1 meter, her potential energy increases by (mass)x(gravitational acceleration)x(1meter), which is the amount of work done.

(This all assumes lifting vertically and no other forces being part of the picture)

5 0

2 years ago

A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric

jeka94

Answer:

a) F₃₁ = 63.0 μN

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

$F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}} = 63.0 \mu N (1)$

Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

$F_{32} = F_{3} - F_{31} = 49.0\mu N - 63.0\mu N = -14.0 \mu N (2)$

Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

$q_{2} = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC (3)$

6 0

2 years ago