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patriot [66]
3 years ago
8

HELP ASAP WITH number 6 plz

Physics
1 answer:
Alika [10]3 years ago
6 0
The answer would be c
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Which statement best define energy
Ksju [112]

Answer:

"The capacity of a system to perform work of any type."

Explanation:

The best statement to describe Energy is:

"The capacity of a system to perform work of any type."

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3 years ago
A 500kg car is driven forward with a thrust force of 1500N. Air resistance and friction acts against the motion of the car with
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2m/s^2, this is because F=ma, meaning a is also equal to F/m. The car applies 1500N in one direction and outside sources apply a total of -500N, meaning the 500kg car is moving forward with a total of 1000N of force. Taking the total 1000N and dividing it by 500kg gives you and acceleration of 2m/s^2. Hope this helps!
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2 years ago
Hooke's law. The distance d when a spring is stretched by a hanging objective varies directly as the weight w of the object. If
Fofino [41]
<span>Hooke's law is F=-kx, which means the elastic force contained by the spring is a product of the distance it stretches and its spring constant, but the direction of the force is opposite that of the displacement. We calculate as follows:

</span><span>(3 kg)(9.8 m/s^2) = -k(-0.38 m)
</span>k =<span> 77.4

</span><span>Then use k to find the new displacement, again using Hooke's law:
(7 kg)(9.8 m/s^2) = -(77.4)x
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7 0
3 years ago
Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m
Phoenix [80]

Answer:

a. t_1=12.5\ s

b. a_2=-13.61\ m.s^{-2}  must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. t_2=2.5714\ s is the time taken to stop after braking

Explanation:

Given:

  • speed of leading car, u_1=25\ m.s^{-1}
  • speed of lagging car, u_{2}=35\ m.s^{-1}
  • distance between the cars, \Delta s=45\ m
  • deceleration of the leading car after braking, a_1=-2\ m.s^{-2}

a.

Time taken by the car to stop:

v_1=u_1+a_1.t_1

where:

v_1=0 , final velocity after braking

t_1= time taken

0=25-2\times t_1

t_1=12.5\ s

b.

using the eq. of motion for the given condition:

v_2^2=u_2^2+2.a_2.\Delta s

where:

v_2= final velocity of the chasing car after braking = 0

a_2= acceleration of the chasing car after braking

0^2=35^2+2\times a_2\times 45

a_2=-13.61\ m.s^{-2} must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

v_2=u_2+a_2.t_2

0=35-13.61\times t_2

t_2=2.5714\ s  is the time taken to stop after braking

7 0
3 years ago
The potential difference across a variable resistor is 11V and the current flowing through it is 0.4A.
klemol [59]
The resistance is 27.5 ohms
5 0
3 years ago
Read 2 more answers
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