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miv72 [106K]
2 years ago
10

What can be defined as the rate at which velocity changes

Physics
2 answers:
siniylev [52]2 years ago
8 0

The rate at which velocity changes can be defined as acceleration or average acceleration.

<u>Explanation:</u>

The acceleration is defined as the rate of change of velocity with respect to time. So, the change in velocity from the initial velocity at different time intervals will help to attain the acceleration attained by the object.

                     \ a =\frac{v_{f}-v_{i}}{\Delta t}

where,

v_{f} \text { and } v_{i} – final and initial velocities, \Delta t– time  

If the change in velocity is constant for same time interval, then the acceleration will be constant for that object. So, an object can attain uniform acceleration when the velocity changes uniformly with time.

taurus [48]2 years ago
4 0

Answer:

acceleration

Explanation:

The rate at which velocity changes is the definition of the physical quantity called acceleration, and it is given by the formula: a=\frac{v_f-v_i}{\Delta t}

where \Delta t is the time that took to change from the initial velocity v_i to the final velocity v_f

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3 years ago
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noname [10]

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3 years ago
A car moves uphill at 40 km/h and then back downhill at 60 km/h. What is the average speed for the round trip?
jok3333 [9.3K]

Answer:

S_a_v_e_r_a_g_e=48km/h

Explanation:

Ok, the average speed can be calculate with the next equation:

S_a_v_e_r_a_g_e=\frac{Total\hspace{3}distance}{Total\hspace{3}time}   (1)

Basically the car cover the same distance "d" two times, but at different speeds, so:

Total\hspace{3}distance=2*d

and the total time would be the time t1 required to go from A to B plus the time t2 required to go back from B to A:

Total\hspace{3}time=t1+t2

From basic physics we know:

t=\frac{d}{S1}

so:

t1=\frac{d}{S1}

t2=\frac{d}{S2}

Using the previous information in equation (1)

S_a_v_e_r_a_g_e=\frac{2*d}{\frac{d}{S1} +\frac{d}{S2} }=\frac{2*d}{\frac{d*S2+d*S1}{S1+S2} }

Factoring:

S_a_v_e_r_a_g_e=\frac{2*S1*S2}{S1+S2}   (2)

Finally, replacing the data in (2)

S_a_v_e_r_a_g_e=\frac{2*40*60}{60+40} =48km/h

5 0
2 years ago
A delivery truck leaves a warehouse and travels 3.20 km east. The truck makes a right turn and travels 2.45 km south to arrive a
boyakko [2]

Explanation:

It is given that,

Displacement of the delivery truck, d_1=3.2\ km (due east)

Then the truck moves, d_2=2.45\ km (due south)

Let d is the magnitude of the truck’s displacement from the warehouse. The net displacement is given by :

d=\sqrt{d_1^2+d_2^2}

d=\sqrt{3.2^2+2.45^2}

d = 4.03 km

Let \theta is the direction of the truck’s displacement from the warehouse from south of east.

\theta=tan^{-1}(\dfrac{2.45}{3.2})

\theta=37.43^{\circ}

So, the magnitude and direction of the truck’s displacement from the warehouse is 4.03 km, 37.4° south of east.

8 0
3 years ago
If 100 J of electrical energy enter the bulb and 5 J of light energy leave the bulb, how many joules of heat energy leave the bu
Novay_Z [31]

As per energy conservation we know that

Energy enter into the bulb = Light energy + Thermal energy

so now we have

energy enter into the bulb = 100 J

Light energy = 5 J

now from above equation we have

100 = 5 + heat

Heat = (100 - 5) J

Heat = 95 J


6 0
3 years ago
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