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3 years ago

11

a car accelerates at a constant rate from 15 m/s to 25 m/s while it travels a distance of 125 m. How long does it take to achiev

e this speed?

1 answer:

atroni [7]3 years ago

6 0

T=Vf-Vi/s
25m/s -15m/s/ 125m
10m/s /125m
=0.08s
I hope it’s correct !

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Even when the head is held erect, as in the figure below, its center of mass is not directly over the principal point of support

alexandr1967 [171]

We are asked to determine the force required by the neck muscle in order to keep the head in equilibrium. To do that we will add the torques produced by the muscle force and the weight of the head. We will use torque in the clockwise direction to be negative, therefore, we have:

$\Sigma T=r_{M\perp}(F_M)-r_{W\perp}(W)$

Since we want to determine the forces when the system is at equilibrium this means that the total sum of torque is zero:

$r_{M\perp}(F_M)-r_{W\perp}(W)=0$

Now, we solve for the force of the muscle. First, we add the torque of the weight to both sides:

$r_{M\perp}(F_M)=r_{W\perp}(W)$

Now, we divide by the distance of the muscle:

$(F_M)=\frac{r_{W\perp}(W)}{r_{M\perp}}$

Now, we substitute the values:

$F_M=\frac{(2.4cm)(50N)}{5.1cm}$

Now, we solve the operations:

$F_M=23.53N$

Therefore, the force exerted by the muscles is 23.53 Newtons.

Part B. To determine the force on the pivot we will add the forces we add the vertical forces:

$\Sigma F_v=F_j-F_M-W$

Since there is no vertical movement the sum of vertical forces is zero:

$F_j-F_M-W=0$

Now, we add the force of the muscle and the weight to both sides to solve for the force on the pivot:

$F_j=F_M+W$

Now, we plug in the values:

$F_j=23.53N+50N$

Solving the operations:

$F_j=73.53N$

Therefore, the force is 73.53 Newtons.

8 0

1 year ago

If the torque required to loosen a nut that holds a wheel on a car has a magnitude of 55 n·m, what force must be exerted at the

erastova [34]

Either 175 N or 157 N depending upon how the value of 48Â° was measured from.
You didn't mention if the angle of 48Â° is from the lug wrench itself, or if it's from the normal to the lug wrench. So I'll solve for both cases and you'll need to select the desired answer.
Since we need a torque of 55 NÂ·m to loosen the nut and our lug wrench is 0.47 m long, that means that we need 55 NÂ·m / 0.47 m = 117 N of usefully applied force in order to loosen the nut. This figure will be used for both possible angles.
Ideally, the force will have a 0Â° degree difference from the normal and 100% of the force will be usefully applied. Any value greater than 0Â° will have the exerted force reduced by the cosine of the angle from the normal. Hence the term "cosine loss".
If the angle of 48Â° is from the normal to the lug wrench, the usefully applied power will be:
U = F*cos(48)
where
U = Useful force
F = Force applied
So solving for F and calculating gives:
U = F*cos(48)
U/cos(48) = F
117 N/0.669130606 = F
174.8537563 N = F
So 175 Newtons of force is required in this situation.
If the 48Â° is from the lug wrench itself, that means that the force is 90Â° - 48Â° = 42Â° from the normal. So doing the calculation again (this time from where we started plugging in values) we get
U/cos(42) = F
117/0.743144825 = F
157.4390294 = F
Or 157 Newtons is required for this case.

6 0

3 years ago

A quarterback throws a football 40 yards in 4 seconds.what is the average speed the football

lions [1.4K]

The answer is 10 yards per second

8 0

3 years ago

Read 2 more answers

Two stereo speakers mounted 4.52 m apart on a wall emit identical in-phase sound waves. You are standing at the opposite wall of

Delicious77 [7]

The answer is option c) 0.9 m

4 0

3 years ago

If I connect an inductor (L) to a capacitor (C), I will get an LC oscillator circuit with some natural frequency omega. If I wer

mart [117]

The new natural frequency would be ω/2.

we know that,

$f = \frac{1}{2 pi \sqrt{LC} }$ = ω. -> equation 1

now, when capacitance is quadrupled,

$f' = \frac{1}{2 pi \sqrt{L ( 4C )} }$

$f' = \frac{1}{2 pi (2)\sqrt{LC} }$ . -> equation 2

substituting value of equation 1 in equation 2 , we get,

$f' = \frac{w}{2}$

Hence, the new natural frequency of the circuit is ω/2.

what do you mean by frequency ?

The resonant frequency for a particular circuit is the frequency at which this equality stands true. Where L is the inductance in henries and C is the capacitance in farads, this is the LC circuit's resonant frequency.

Learn more about frequency here:-

brainly.com/question/12530980

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2 years ago