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3 years ago

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A horizontal force in used to pull a 5 kilogram cart at a constant speed of 5 meters per second across the floor as shown in the

diagram. if the force of friction between the cart and the floor is 10 newtons, the magnitude of the horizontal force along the handle of the cart is?
a. 5N
b. 10 N
c. 25 N
d. 50 N

1 answer:

ddd [48]3 years ago

5 0

A cart is pulled by horizontal force such that it moves with constant velocity

So here since velocity is constant we can say that its acceleration will be ZERO

now here for zero acceleration we can say

$F_{net} = 0$

So here we will have

$F_{net} = F_{ap} - F_f = 0$

here we know that

$F_f = 10 N$

so we will have

$F_{ap} - 10 = 0$

$F_{ap} = 10 N$

so here applied force on handle will be

<em>b. 10 N</em>

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Can someone help me?!!!!!

german

<h2>Hello!</h2>

The answer is:

The first option, the walker traveled 360m more than the actual distance between the start and the end points.

Why?

Since each block is 180 m long, we need to calculate the vertical and the horizontal distance, in order to calculate how farther did the travel walk between the start and the end points (displacement).

So, calculating we have:

Traveler:

$Distance=NorthCoveredDistance+EastCoveredDistance$

$Distance=4*180m+3*180m=720m+540m=1260m$

Actual distance between the start and the end point (displacement):

$ActualDistance=\sqrt{NorthDistance+EastDistance}\\\\ActualDistance=\sqrt{NorthDistance^{2} +EastDistance^{2}}\\\\ActualDistance=\sqrt{(720m)^{2} +(540m)^{2}}\\\\ActualDistance=\sqrt{518400m^{2} +291600m^{2}}\\\\ActualDistance=\sqrt{810000m^{2}}=900m$

Now, to calculate how much farter did the traveler walk, we need to use the following equation:

$DistanceDifference=WalkerCoveredDistance-ActualDistance\\\\DistanceDifference=1260m-900m=360m$

Therefore, we have that distance differnce between the distance covered by the walker and the actual distance is 360m.

Hence, we have that the walker traveled 360m more than the actual distance between the start point and the end point.

Have a nice day!

3 0

2 years ago

Allen and Jason are chucking a speaker around. On one particular throw, Allen throws the speaker, which is playing a pure tone o

Sholpan [36]

Answer:

Explanation:

We shall apply Doppler's effect of sound .

speaker is the source , Jason is the observer . Source is moving at 10 m /s , observer is moving at 6 m/s .

apparent frequency = $f_o\times\frac{V+v_o}{ V-v_s}$

V is velocity of sound , v₀ is velocity of observer and v_s is velocity of source and f_o is real frequency of source .

Here V = 340 m/s , v₀ is 6 m/s , v_s is 10 m/s . f_o = f

apparent frequency = $f\times \frac{340+6}{340-10}$

= $f\times \frac{346}{330}$

So m = 346 , n = 330 .

8 0

3 years ago

A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.1 rad/s in 2.99 s.(a) fi

elena55 [62]

The angular acceleration of a rotating object is given by
$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$
where
$\omega_f$ is the final angular speed of the object
$\omega_i$ is its initial angular speed
$\Delta t$ is the time taken to accelerate

For the wheel in our problem, $\omega_f=11.1 rad/s$ , $\omega_i = 0$ and $\Delta t=2.99 s$ , so its angular acceleration is
$\alpha= \frac{11.1 rad/s-0}{2.99 s}=3.71 rad/s^2$

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2 years ago

What do the peaks represent

monitta

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3 years ago

A city planner is working on the redesign of a hilly portion of a city. An important consideration is how steep the roads can be

Artyom0805 [142]

Explanation:

It is known that relation between force and acceleration is as follows.

F = $m \times a$

I is given that, mass is 1090 kg and acceleration is 21 m/s. Therefore, we will calculate force as follows.

F = $m \times a$

= $1090 \times (\frac{21}{16})$

= 1430.625 N

Also, it is known that

$sin(\theta) = \frac{\text{Force car can exert}}{\text{Force gravity pulls car}}$

$sin(\theta) = \frac{1430.625 N}{(1090 \times 9.8) N}$

$\theta$ = 7.70 degrees

Thus, we can conclude that the maximum steepness for the car to still be able to accelerate is 7.70 degrees.

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3 years ago