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2 years ago

6. Can an object accelerate, but not speed up or slow down? Explain:

Physics

1 answer:

FromTheMoon [43]2 years ago

5 0

Answer:

yes, though the speed may stay constant, the direction will change. so for example, you're going 70mph on the freeway, but you have to take the exit on your right (the exit continues on to a different freeway), you're not going to speed up or slow down, you'll change your direction which is still accelerating.

Explanation:

Credit goes to @naeAF

Hope this helps :))

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A car starts from rest and is moving at 60.0 m/s after 7.50 s. What is the car's average acceleration?

Scrat [10]

Answer:

${ \boxed{ \bold{ \sf{Acceleration \: ( \: a) = 8 \: m/ {s \: }^{2} }}}}$

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♨ Question :

A car starts from rest and is moving at 60.0 m/s after 7.50 s. What is the car's average acceleration ?

♨ $\underbrace{ \sf{Required \: Answer : }}$

☄ Given :

Initial velocity ( u ) = 0
Final velocity ( v ) = 60.0 m/s
Time ( t ) = 7.50 s

☄ To find :

Acceleration ( a )

✒ We know ,

$\boxed{ \underline{ \bold{ \sf{Acceleration \: ( \: a) = \frac{Final velocity ( v ) - Initial velocity ( u)}{t} }}}}$

Substitute the values and solve for a.

➛ $\sf{a = \frac{60.0 - 0}{7.50}}$

➛ $\sf{a = \frac{60.0}{7.50}}$

➛ $\boxed{ \boxed{ \sf{a = 8 \: m/ {s \: }^{2} }}}$

---------------------------------------------------------------

✑ Additional Info :

When a certain object comes in motion from rest , in the case , initial velocity ( u ) = 0
When a moving object comes in rest , in the case , final velocity ( v ) = 0
If the object is moving with uniform velocity , in the case , u = v.
If any object is thrown vertically upwards in the case , a = -g
When an object is falling from certain height , in the case , final velocity at maximum height ( v ) = 0.

Hope I helped!

Have a wonderful time ツ

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2 years ago

A swimming pool has the shape of a box with a base that measures 30 m by 12 m and a uniform depth of 2.2 m. How much work is req

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Answer:

853776 J

Explanation:

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E = Wh = mgh

Since water mass is a body of water we can treat it as the product of density 1000kg/m3 and volume, which is the product of base area and uniform height h

$m = \rho V = \rho \int\limits^{2.2}_0 {A} \, dh$

Therefore:

$E = mgh = g\rho A\int\limits^{2.2}_0 {h} \, dh\\E = 9.8*1000*30*12[h^2/2]^{2.2}_0 = 1764000(2.2^2 - 0^2) = 853776 J$

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The force between two point-charges of 0.040 C and 0.060 C separated by certain distance is 2.2 x 10^5 N. what is the distance b

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The upward velocity of a 2540kg rocket is v(t)=At + Bt2. At t=0 a=1.50m/s2. The rocket takes off and one second afterwards v=2.0

Alexeev081 [22]

Answer:

The value of A is 1.5m/s^2 and B is 0.5m/s^³

Explanation:

The mass of the rocket = 2540 kg.

Given velocity, v(t)=At + Bt^2

Given t =0

a= 1.50 m/s^2

Now, velocity V(t) = A*t + B*t²

If, V(0) = 0, V(1) = 2

a(t) = dV/dt = A+2B × t

a(0) = 1.5m/s^²

1.5m/s^² = A + 2B × 0

A = 1.5m/s^2

now,

V(1) = 2 = A× 1 + B× 1^²

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Now Check V(t) = A× t + B × t^²

So, V(1) = A× (1s) + B× (1s)^² = 1.5m/s^² × 1s + 0.5m/s^³ × (1s)^² = 1.5m/s + 0.5m/s = 2m/s

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