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alexira [117]
3 years ago
9

The front end of the car is crushing and absorbing ___ which slows down the rest of the car

Physics
1 answer:
Luda [366]3 years ago
8 0

The answer is energy


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As you travel from Detroit in a certain direction, the outside temperature, T (in degrees), depends on your distance, d (in mile
Ber [7]

Answer:

a)\Delta T= 100^{\circ}C

b)\bigtriangledown T=1^{\circ}C.mile^{-1}

c)\bigtriangledown T_4=1^{\circ}C.mile^{-1}

d)\bigtriangledown T_4=1^{\circ}C.mile^{-1}

Explanation:

Given is the data of variation of temperature with respect to the distance traveled:

Temperature T as a function of distance d:

T=(d+30) ^{\circ}C...................................(1)

(a)

Total change in temperature from the start till the end of the journey:

\Delta T= T_f-T_i..............................(2)

where:

T_f= final temperature

T_i= initial temperature

∵In the start of the journey d = 0 miles & at the end of the journey d = 100 miles.

So, correspondingly we have the eq. (2) & (1) as:

\Delta T= (100+30)-(0+30)

\Delta T= 100^{\circ}C

(b)

Now, the average rate of change of the temperature, with respect to distance, from the beginning of the trip to the end of the trip be calculated as:

\bigtriangledown T=\frac{\Delta T}{\Delta d}......................(3)

where:

\Delta d= change in distance

\bigtriangledown T=change in temperature with respect to distance

putting the respective values in eq. (3)

\bigtriangledown T=\frac{100}{100}

\bigtriangledown T=1^{\circ}C.mile^{-1}

(c)

comparing the given function of the temperature with the general equation of  a straight line:

y=m.x+c

We find that we have the slope of the equation as 1 throughout the journey and therefore the rate of change in temperature with respect to distance remains constant.

\bigtriangledown T_4=1^{\circ}C.mile^{-1}

(d)

comparing the given function of the temperature with the general equation of  a straight line:

y=m.x+c

We find that we have the slope of the equation as 1 throughout the journey and therefore the rate of change in temperature with respect to distance remains constant.

\bigtriangledown T_4=1^{\circ}C.mile^{-1}

4 0
3 years ago
Objects A and B are brought close to each other. Object A will soon become positively charged. Identify the charge that must tra
Rzqust [24]
B because that’s why they are going out of the circle
4 0
3 years ago
A strong person and a weak person are trying to carry a ladder.How should they carry it in such a way that the weak person feels
egoroff_w [7]

Answer:

The stronger person holds it closer to them.

Explanation:

if the stronger person were to hold the ladder closer to them, they would be holding more of the weight thus making the weaker person have less weight. this is because of the center of gravity being closer to the stronger person.

8 0
3 years ago
Read 2 more answers
5. A car accelerates from rest down a highway at 6 m/s2 for 5 seconds. What distance did the car travel during acceleration?
Ivenika [448]

Answer:75m

Explanation: simple kinematics  . xo=0 , vo=0

x=xo+vo*t +at*t/2 = 6*5*5/2 =75m

6 0
3 years ago
Calculate the potential at the center of curvature of the arc if the potential is assumed to be zero at infinity.
yan [13]

the potential at the center of curvature of the arc = v = Q ∕ (4πε∘a) or \frac{kQ}{a}

ATQ,

We have density of charge,

λ = \frac{Q}{L}

Where L is the rod's length, in this case the semicircle's length L = πr

Q is the charge on the rod

The potential created at the center by an differential element of charge is:

k = \frac{dQ}{r}

where k is the coulomb's constant

r is the distance from dQ to center of the circle

v = ∫ \frac{K dQ}{a} , Where a = radius, k = 1 / 4πε∘

v =\frac{kQ}{a} or Q ∕ (4πε∘a)

To learn more about potential from the given link

brainly.com/question/25923373

#SPJ4

6 0
1 year ago
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