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3 years ago

An object accelerates from rest to 85m/s over a distance of 36m. What acceleration did it experience?

1 answer:

6 0

Suvat
we have s, u, v and we want a
the suvat equation with these values in is: v^2 = u^2 - 2as
so a = (-v^2 + u^2)/-2s
plug numbers in
a = (-85^2 + 0^2)/-2*36 = 7225/72 = 100.3... ms^-2

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A bicyclist starts at rest and speeds up to 30 m/s while accelerating at 4 m/s^2. Determine the distance traveled.

raketka [301]

Answer:

Distance, d = 112.5 meters

Explanation:

Initially, the bicyclist is at rest, u = 0

Final speed of the bicyclist, v = 30 m/s

Acceleration of the bicycle, $a=4\ m/s^2$

Let s is the distance travelled by the bicyclist. The third equation of motion is given as :

$v^2-u^2=2as$

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{(30)^2}{2\times 4}$

s = 112.5 meters

So, the distance travelled by the bicyclist is 112.5 meters. Hence, this is the required solution.

6 0

2 years ago

A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike

Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g= $9.50\times 10^{3} kg$

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

$mu+ MU=mv+ MV$

Substitute the values

$9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V$

$3.61i=2.375j+0.150V$

$3.61 i-2.375j=0.150V$

$V=\frac{1}{0.150}(3.61 i-2.375j)$

$V=24.07i-15.83j$

Magnitude of velocity of stone

= $\sqrt{(24.07)^2+(-15.83)^2}$

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

$\theta=tan^{-1}(\frac{y}{x})$

= $tan^{-1}(\frac{-15.83}{24.07})$

$\theta=tan^{-1}(-0.657)$

=33.3 degree below the horizontal

(b)

Initial kinetic energy

$K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2$

$K_i=685.9 J$

Final kinetic energy

$K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2$

= $\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2$

$K_f=359.12 J$

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

5 0

3 years ago

10. A worker uses a pulley system to raise a 24.0 kg carton 16.5 m. A force of 129 N is exerted and the rope is pulled 33.0 m. a

lozanna [386]

Answer: Machanical advantage of the machine is 1.86

Explanation: Machanical advantage of a machine is the ratio of the Force to overcome which is the load in this case 24kg * 10= 240N to the force exerted(Effort) to overcome the load in this case 129N.

So, we have

MA = load/effort

= 240N/129N

= 1.86.

5 0

2 years ago

Explain how machines can be useful if the output is always less than the input work

Anni [7]

Because: Some of the work done by the machine is used to overcome the friction created by the use of the machine. ... Work output can never be greater than work input. Machines allow force to be applied over a greater distance, which means that less force will be needed for the same amount of work.

6 0

3 years ago

A 22.0 kg bucket of concrete is connected over a very light frictionless pulley to a 375 N box on the roof of a building as show

Veronika [31]

Answer:

vf = 3.27[m/s]

Explanation:

In order to solve this problem we must analyze each body individually and find the respective equations. The free body diagram of each body (box and bucket) should be made, in the attached image we can see the free body diagrams and the respective equations.

With the first free body diagram, we determine that the tension T should be equal to the product of the mass of the box by the acceleration of this.

With the second free body diagram we determine another equation that relates the tension to the acceleration of the bucket and the mass of the bucket.

Then we equalize the two stress equations and we can clear the acceleration.

a = 3.58 [m/s^2]

As we know that the bucket descends 1.5 [m], this same distance is traveled by the box, as they are connected by the same rope.

$x = \frac{1}{2} *a*t^{2}\\1.5 = \frac{1}{2}*(3.58) *t^{2} \\t = 0.91 [s]$

And the speed can be calculated as follows:

$v_{f}=v_{o}+a*t\\v_{f}=0+(3.58*0.915)\\v_{f}= 3.27[m/s]$

7 0

3 years ago