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3 years ago

A racing car increases its speed from 10 m/s to 50 m/s over a distance of 60 m.

1 answer:

Nookie1986 [14]3 years ago

5 0

We can solve this by using the equation

S = ut + 1/2 at^2
Where S is distance
And u is initial velocity
And a is acceleration

a = v-u/t

So
S = ut + 1/2 x v-u/t x t^2
60 = 10t + 1/2 x 50 - 1/2 x 10 x t
60 = 10t + 25 - 5 x t
60= 10t + 20t
60 = 30t
60/30 = t
2 equals t

So it takes 2 seconds

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An archer shoots an arrow 75 m distant target; the bull's-eye of the target is at the same height as the release height of the a

Liono4ka [1.6K]

Answer:

$16.25^{\circ}$

Explanation:

R = Horizontal range of projectile = 75 m

v = Velocity of projectile = 37 m/s

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

Horizontal range is given by

$R=\dfrac{v^2\sin2\theta}{g}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{Rg}{v^2}}{2}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{75\times 9.81}{37^2}}{2}\\\Rightarrow \theta=16.25^{\circ}$

The angle at which the arrow is to be released is $16.25^{\circ}$ .

4 0

3 years ago

) Electric charge is uniformly distributed inside a nonconducting sphere of radius 0.30 m. the electric field at a point P, whic

aleksandr82 [10.1K]

Answer:

Explanation:

Point P is situated outside sphere . For a non conducting sphere , electric field outside sphere is given by the relation

E = k Q / d²

15000 = 9 x 10⁹ x Q / .5²

Q = 416.67 X 10⁻⁹

Maximum field will be when d = r ( radius )

maximum E = kQ / r²

= 9 x 10⁹ x 416.67 X 10⁻⁹ / .3²

= 41667 N / C

8 0

3 years ago

A 2.00 kg block situated on a rough incline is connected to aspring

garri49 [273]

Figure P5.76 is missing from this problem, but I found a similar problem whose figure I uploaded with this answer. The angle of the incline might be different though, but the procedure to follow is the same.

Answer:

$\mu_{k}=0.0708$

Explanation:

In order to solve this problem we must first do a drawing of the situation so we can analyze it better. (See attached picture)

Once we have the drawing of the problem, we can go ahead and draw a free body diagram, which will help us determine what forces are acting upon the object. There are different approaches we can take to solve this problem, but I will use an energy balance to do so.

So first we do a sum of forces on the y-axis. This is for us to find what the Normal force is, which will be used to find the force of friction, so we get:

$\sum F_{y}=0$

by looking at the free body diagram we get the sum of forces to be:

$N-W_{y}=0$

when solving for the normal force we get that:

$N=W_{y}$

we also know from the free body diagram that:

$W_{y}=Wcos \theta$

and

$W_{y}=mgcos \theta$

So:

$N=mgcos \theta$

we can also determine the height of the block at the time it is released by analyzing the triangle find in the uploaded figure,, so we get that:

$sin(37^{o})=\frac{h}{21.4cm}$

so:

h=21.4cm sin(37°)

h=12.88cm

Once we got this, we can go ahead and do an energy balance on the system, so we get that:

$U_{0}+K_{0}+E_{s0}=U_{f}+K_{f}+W_{f}+E_{sf}$

where:

U=potential energy

K=Kinetic energy

$W_{f}$ =Work of friction

$E_{s}$ =Potential energy of the spring

Since the object is released from rest we know the initial kinetic energy is zero, just like the initial potential energy of the spring since when the block is released, the spring is unstretched. We also know the final potential energy of the block is zero because it reached its lowest point, while the kinetic energy of the block is also zero because it came to rest at that point. This simplifies our energy balance so we get:

$U_{0}=W_{f}+E_{sf}$

we can now determine each part of the equation so:

$U_{0}=mgh$

$W_{f}=fx$

we know that friction is given gy the equation:

$f=\mu_{k}N$

and that

$N=mgcos \theta$

so:

$f=\mu_{k}mgcos \theta$

and

$W_{f}=\mu_{k}mgxcos \theta$

and finally:

$E_{s}=\frac{1}{2}kx^{2}$

Once we got all these equations we can substitute them into our balance of energy, so we get:

$mgh=\mu_{k}mgx cos \theta + \frac{1}{2}kx^{2}$

we can now solve this for $\mu_{k}$ so we get:

$mgh - \frac{1}{2}kx^{2}=\mu_{k}mgx cos \theta$

and:

$\mu_{k}=\frac{2mgh-kx^{2}}{2mgxcos\theta}$

now we can substitute the given data, make sure to use the correct units:

$\mu_{k}=\frac{2(2kg)(9.81m/s^{2})(12.88x10^{-2}m)-(100N/m)(21.4x10^{-2})^{2}}{2(2kg)(9.81m/s^{2})(21.4x10^{-2})cos(37^{o})}$

which solves to:

$\mu_{k}=0.0708$

5 0

4 years ago

According to the impulse--momentum equation Ft = change in mv, a bungee jumper in fall has momentum which is reduced by the forc

Dima020 [189]

Answer:

The answer is a. Jumper

Explanation:

The change in the momentum is represented by the equation

$Ft=m*v$

Acceleration of any body is calculated by the product of the mass of the body and with the velocity the same mass is moving.

Since it is mentioned in the statement that m is the mass of the jumper then velocity will also be of the jumper to calculate the change in the momentum.

If the mass of bugle cord is taken, then its velocity will be taken for the equation.

6 0

4 years ago

two individuals start the training program at the same time, but one is able to grow muscle faster an larger than other. what is

Pavel [41]

One individual was born with more muscle fibers.

4 0

3 years ago