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3 years ago

5

A car is traveling down the highway cruise control at 50m/s (this means its is traveling with a constant velocity NOT accelerati

ng). If it does this for 5 minutes (300s), how far does it travel?

1 answer:

lawyer [7]3 years ago

3 0

Given parameters:

Speed of car = 50m/s

Time of travel = 5minutes (300s)

Unknown;

Distance of travel = ?

To solve this problem, we need to understand how speed relates with distance and time.

Speed is a physical quantity. It is the distance traveled divided by time taken.

Speed = $\frac{distance}{time}$

So,

Distance = speed x time

Input the parameters and solve for the distance;

Distance = 50m/s x 300s

Distance = 15000m

The distance covered during this time interval and speed is 15000m

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A 67.0 kg crate is being raised by means of a rope. Its upward acceleration is 3.50 m/s2. What is the force exerted by the rope

serious [3.7K]

<span>A 67.0 kg crate is being raised by means of a rope. Its upward acceleration is 3.50 m/s2. What is the force exerted by the rope on the crate?

</span>Newton's Second Law<span> of Motion states, “The force acting on an object is equal to the mass of that object times its acceleration.” We calculate as follows:
</span>
F = ma = 67.0 kg (3.50 m/s^2) = 234.5 J

7 0

3 years ago

Which equations can be used to solve for acceleration? Select four options.

tatiyna

Explanation:

Acceleration of an object is calculated by finding the change in its velocity divided by time taken.

If $v_i$ is initial velocity, $v_f$ is final velocity and t is time taken. Then the acceleration of the object is given by :

$a=\dfrac{v_f-v_i}{t}\\\\at=v_f-v_i\\\\v_f=v_i+at$ .....(1)

So, the above equation is used to find acceleration. It is called the first equation of motion. After rearranging equation (1), the correct options are :

$t=\dfrac{\Delta v}{a}$

$v_i=v_f-at$

$v_f=v_i+at$

$a=\dfrac{\Delta v}{t}$

7 0

3 years ago

Read 2 more answers

Two carts, one of mass 2m and one of mass m, approach each other with the same speed, v. When the carts collide, they hook toget

Nikitich [7]

Answer:

<em>Second option</em>

Explanation:

<u>Linear Momentum</u>

The linear momentum of an object of mass m and speed v is

P=mv

If two or more objects are interacting in the same axis, the total momentum is

$P_t=m_1v_1+m_2v_2+...$

Where the speeds must be signed according to a fixed reference

The images show a cart of mass 2m moves to the left with speed v since our reference is positive to the right

$P_1=-2mv$

The second cart of mass m goes to the right at a speed v

$P_2=mv$

The total momentum before the impact is

$P_t=-2mv+mv=-mv$

The total momentum after the collision is negative, both carts will join and go to the left side

The first option shows both carts with the same momentum before the collision and therefore, zero momentum after. It's not correct as we have already proven

The third option shows the 2m cart has a positive greater momentum than the other one. We have proven the 2m car has negative momentum. This option is not correct either

The fourth option shows the two carts keep separated after the collision, which contradicts the condition of the question regarding "they hook together".

The second option is the correct one because the mass $m_2$ has a negative momentum and then the sum of both masses keeps being negative

3 0

4 years ago

Read 2 more answers

Unpolarized light passes through a vertical polarizing filter, emerging with an intensity I0. The light then passes through a ho

antiseptic1488 [7]

To solve this problem it is necessary to apply the concepts related to Malus' law. Malus' law indicates that the intensity of a linearly polarized ray of light that passes through a perfect analyzer with a vertical optical axis is equivalent to:

$I = I_0 cos^2(\theta)$

$I_0 =$ Indicates the intensity of the light before passing through the Polarizer,

I = The resulting intensity, and

$\theta$ = Indicates the angle between the axis of the analyzer and the polarization axis of the incident light.

There is 3 polarizer, then

For the exit of the first polarizer we have that the intensity is,

$I_2 = I_0cos^2(45)$

For the third polarizer then we have,

$I_3 = I_2 cos^{2}(45)$

Replacing with the first equation,

$I_3 = I_0cos^2(45)cos^{2}(45)$

$I_3 = I_0 (\frac{1}{2})(\frac{1}{2})$

$I_3 = I_0 \frac{1}{4}$

Therefore the transmitted intensity now is $\frac{1}{4}$ of the initial intensity.

5 0

4 years ago

Pages 64 to 65 of your reading material trace an extended problem that involes lifting a bag of sugar up to a shelf at first it

nadezda [96]

Answer:

d) the amount of work is the same whether the bag is moved all at once or in two stages, provided the total height lifted is the same in either case.

Explanation:

While moving the bag to the shelf in one shot we can say that the total work done is given as

$W = mg(2H)$

here we know that

2H = total height raised by the bag

now when we raise the bag to first shelf and then move it to next shelf

then we will have

$W = W_1 + W_2[tex][tex]W = mgH + mgH$

$W = 2mgH$

so the correct answer will be

d) the amount of work is the same whether the bag is moved all at once or in two stages, provided the total height lifted is the same in either case.

7 0

3 years ago