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3 years ago

A car accelerates from rest (v. 0 m/s) with a constant acceleration of

Physics

1 answer:

oksano4ka [1.4K]3 years ago

4 0

Heya!!

For calculate final velocity, lets applicate formula

$\boxed{V=V_o+a*t}$

<u>Δ Being Δ</u>

V = Final Velocity = ?

Vo = Initial velocity = 0 m/s

a = Aceleration = 5 m/s²

t = Time = 12 s

⇒ Let's replace according the formula:

$\boxed{V=0\ m/s +5\ m/s*12\ s}$

⇒ Resolving

$\boxed{V=60\ m/s}$

Result:

The velocity after 10 sec is <u>60 meters per second (m/s)</u>

Good Luck!!

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4 years ago

A motorcycle accelerates uniformly from rest and reaches a linear speed of 24.8 m/s in a time of 9.87 s. The radius of each tire

Vinvika [58]

Answer:

8.756 rad/s²

Explanation:

Given that:

A motorcycle accelerates uniformly from rest, then initial velocity v_i = 0 m/s

It final velocity v_f = 24.8 m/s

time (t) = 9.87 s

radius (r) of each tire = 0.287 m

Firstly; the linear acceleration of the motor cycle is determined as follows:

$a_T$ =(V_f - v_i)/t

=(24.8-0)/9.87

=2.513 m/s²

Then; the magnitude of angular acceleration

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=2.513/0.287

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6 0

3 years ago

Read 2 more answers

A 70 kg student now sits in the chair and the same 35 N force is applied to the chair. What is the chair's new acceleration?

amid [387]

Answer:

0.5 m/s²

Explanation:

Applying,

F = ma............... Equation 1

Where F = Force applied to the chair, m = mass of the student, a = acceleration of the chair

make a the subject of the equation

a = F/m............ Equation 2

From the question,

Given: F = 35 N, m = 70 kg

Substitute these values into equation 2

a = 35/70

a = 0.5 m/s²

Hence the new acceleration of the chair is 0.5 m/s²

5 0

3 years ago

A kangaroo jumps up with an initial velocity of 36 feet persecond from the ground (assume its starting height is 0 feet).Use the

kkurt [141]

Given

Initial velocity:

36 ft/s

Initial height:

0 ft

Vertical motion model:

h(t) = -16t^2 + ut + s

v = initial velocity

s = is the height

Procedure

We are going to use the model provided for the vertical motion.

$\begin{gathered} h(t)=-16t^2+36t+0 \\ h(t)=-16t^2+36t \end{gathered}$

We know that at the maximum height the final velocity is 0.

Then we will use the following expression to calculate the maximum height:

$\begin{gathered} v^2_f=v^2_o-2ah_{\max } \\ 0=v^2_o-2ah_{\max } \\ 2ah_{\max }=v^2_o \\ h_{\max }=\frac{v^2_o}{2a} \\ h_{\max }=\frac{(36ft/s)^2}{2\cdot32ft/s^2} \\ h_{\max }=20.25\text{ ft} \end{gathered}$

Now for time:

$\begin{gathered} 20.25=-16t^2+36t \\ 16t^2-36t+20.25=0 \end{gathered}$

Solving for t,

$\begin{gathered} t_1=2.25 \\ t_2=0 \end{gathered}$

The total time the kangaroo takes in the air is 2.3s.

3 0

1 year ago