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Reil [10]
3 years ago
5

What force is necessary to accelerate a 1050 kg car at a rate of 10 m/s2?

Physics
1 answer:
skelet666 [1.2K]3 years ago
5 0
M=2000
A=5 m/s^2
F=(2000kg)
F=10000N
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What is an example of a renewable natural resource?
marin [14]

Answer:

Biomass energy, Wind Energy, Hydro-power, Solar energy are some of the most utilized naturally occuring renewable resources.

6 0
3 years ago
The acceleration due to gravity on the moon is about 1/6 of the acceleration due to gravity on the earth. A net force F acts hor
uysha [10]

Answer:

c.a_m

Explanation:

We are given that

Acceleration due to gravity on the moon=a_m

Acceleration due to gravity on the earth=a_e

g_m=\frac{1}{6}g_e

Net force due to am on an object on moon=F_{net}=ma_m

There is no friction and no drag force and there is no gravity involved

Then, the force acting on an object on earth=F=ma_e

F=F_{net}(given)

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3 0
3 years ago
How can we produce energy by turning a turbine?​
maw [93]
Wind turbines work on a simple principle: instead of using electricity to make wind—like a fan—wind turbines use wind to make electricity. Wind turns the propeller-like blades of a turbine around a rotor, which spins a generator, which creates electricity.
7 0
4 years ago
Assume a rectangular strip of a material with an electron density of n=5.8x1020 cm-3. The strip is 8 mm wide and 0.8 mm thick an
vampirchik [111]

Answer: I = 111.69 pA

Explanation: The hall effect is all about the fact that when a semiconductor is placed perpendicularly to a magnetic field, a voltage is generated which could be measured at right angle to the current path. This voltage is known as the hall voltage.

The hall voltage of a semiconductor sensor is given below as

V = I×B/qnd

Where V = hall voltage = 1.5mV =1.5/1000=0.0015V

I = current =?,

n= concentration of charge (electron density) = 5.8×10^20cm^-3 = 5.8×10^20/(100)³ = 5.8×10^14 m^-3

q = magnitude of an electronic charge=1.609×10^-19c

B = strength of magnetic field = 5T

d = thickness of sensor = 0.8mm = 0.0008m

By slotting in the parameters, we have that

0.0015 = I × 5/5.8×10^14 × 1.609×10^-19×0.0008

0.0015 = I×5/7.446×10^-8

I = (0.0015 × 7.446×10^-8)/5

I = 111.69*10^(-12)

I = 111.69 pA

3 0
3 years ago
The box leaves position x=0x=0 with speed v0v0. The box is slowed by a constant frictional force until it comes to rest at posit
const2013 [10]

Answer:

fr = ½ m v₀²/x

Explanation:

This exercise the body must be on a ramp so that a component of the weight is counteracted by the friction force.

The best way to solve this exercise is to use the energy work theorem

            W = ΔK

Where work is defined as the product of force by distance

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The angle is because the friction force opposes the movement

          Δk =K_{f} –K₀

          ΔK = 0 - ½ m v₀²

We substitute

         - fr x = - ½ m v₀²      

           fr = ½ m v₀²/x

8 0
4 years ago
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