All categories

2 years ago

A sports car can move 100.0 m in the first 4.5 s of constant acceleration.

Physics

2 answers:

Inessa [10]2 years ago

6 0

The car acceleration is 22.2 repeating meters per second squared

Troyanec [42]2 years ago

4 0

Answer:

Acceleration of the car, $a=9.87\ m/s^2$

Explanation:

Initial sped of the car, u = 0

Distance covered by the car, d = 100 m

Time taken, t = 4.5 s

Let a is the acceleration of the car. It can be calculated using second equation of kinematics as :

$d=ut+\dfrac{1}{2}at^2$

$d=\dfrac{1}{2}at^2$

$a=\dfrac{2d}{t^2}$

$a=\dfrac{2\times 100}{(4.5)^2}$

$a=9.87\ m/s^2$

So, the acceleration of the car is $9.87\ m/s^2$ . Hence, this is the required solution.

You might be interested in

A(n) 71.1 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 70.2 m away from the shuttle

denpristay [2]

Answer:

10.347 minutes.

Explanation:

According to F = ma, she exerts force on camera of the magnitude

F = 0.67Kg*12m/ $s^{2}$ = 8.04N, assuming it took her one second to accelerate camera to 12m/s, then by newtons third law, which says every action has equal and opposite reaction , the camera exerts the same amount of force on the astronaut which gives her acceleration of a = $\frac{8.04N}{70.2Kg} = 0.1130801680m/s^2$ .

and velocity of V = 0.1130801680m/s.

at this velocity , the astronaut has to cover the distance of 70.2 meters, it will take her 620.7985075s = 10.347 min to get to the shuttle (using S = vt).

4 0

2 years ago

What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then

Darya [45]

Answer:

$y = -8.37 cm$

Explanation:

As we know that the equation of SHM is given as

$y = A cos(\omega t)$

here we know that

$\omega = \frac{2\pi}{T}$

here we have

$T = 0.66 s$

now we have

$\omega = \frac{2\pi}{0.66}$

$\omega = 3\pi$

now we have

$y = (8.8 cm) cos(3\pi t)$

now at t = 2.3 s we have

$y = (8.8 cm) cos(3\pi \times 2.3)$

$y = -8.37 cm$

6 0

2 years ago

Read 2 more answers

The position of an object that is oscillating on an ideal spring is given by the equation x=(12.3cm)cos[(1.26s−1)t]. (a) at time

Natali5045456 [20]

x=((12.3/100)m)cos[(1.26s^−1)t]
v= dx/dt = -((12.3/100)*1.26)sin[(1.26s^−1)t]
v=-((12.3/100)*1.26)sin[(1.26s^−1)t]=-((12.3/100)*1.26)sin[(1.26s^−1)*(0.815)]
v= -0.13261622 m/s
the object moving at 0.13 m/s at time t=0.815 s

6 0

2 years ago

The pail is rotated at a constant rate so it has the minimum speed at all points along its circular path. The water has mass m.

Alisiya [41]

Answer:

Explanation:

The pail is rotated at a constant rate in vertical circular path so it has the minimum speed at all points along its circular path . That means at top position the velocity is almost zero. In that case the centripetal force at top position will be provided by its weight or

mg = mv² / r ( r is radius of vertical circular path )

v = √ rg

At the bottom position its velocity will be increased due to loss of potential energy

so 1/2 m V² = 1/2 m v² + mg x 2r

V =√ 5 gr

If R be the reaction force at the bottom by bottom of pail

R - mg = mV² / r

R = mg +mV² / r

= mg + m x 5gr / r

R = 6mg

This is the magnitude of the force exerted by the water on the bottom of the pail .

7 0

2 years ago

An object of mass 24kg is accelerated up a frictionless place incline at an angle of 37° with horizontal by a constant force, st

RoseWind [281]

a) Average power: 1425 W

b) Instantaneous power at 3.0 sec: 2850 W

Explanation:

The motion of the object along the ramp is a uniformly accelerated motion (because the force applied is constant), so we can use the suvat equation

$s=ut+\frac{1}{2}at^2$

where

s = 18 m is the displacement along the ramp

u = 0 is the initial velocity

t = 3.0 s is the time taken

a is the acceleration of the object along the ramp

Solving for a,

$a=\frac{2s}{t^2}=\frac{2(18)}{(3.0)^2}=4 m/s^2$

Now we can apply Newton's second law to find the net force on the object:

$F=ma=(24 kg)(4 m/s^2)=96 N$

This net force is the resultant of the applied force forward ( $F_a$ ) and the component of the weight acting backward ( $mg sin \theta$ ), so we can find what is the applied force: