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Sedaia [141]
3 years ago
12

A truck accelerates from a stop to 27m/sec in 9 minutes. What was the trucks acceleration?

Physics
1 answer:
olchik [2.2K]3 years ago
5 0
The acceleration is 3 m/s per minute, or 0.05 m/s per second.
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On the sonometer shown below, a horizontal cord of length 5 m has a mass of 1.45 g. When the cord was plucked the wave produced
Korolek [52]

Answer:

(a) T = 0.015 N

(b) M = 1.53 x 10⁻³ kg = 1.53 g

Explanation:

(a) T = 0.015 N

First, we will find the speed of waves:

v =f\lambda

where,

v = speed of wave = ?

f = frequency = 120 Hz

λ = wavelength = 6 cm = 0.06 m

Therefore,

v = (120 Hz)(0.06 m)

v = 7.2 m/s

Now, we will find the linear mass density of the coil:

\mu = \frac{m}{l}

where,

μ = linear mass density = ?

m = mass = 1.45 g = 1.45 x 10⁻³ kg

l = length = 5 m

Thereforre,

\mu = \frac{1.45\ x\ 10^{-3}\ kg}{5\ m}\\\\\mu = 2.9\ x\ 10^{-4}\ kg/m

Now, for the tension we use the formula:

v = \sqrt{\frac{T}{\mu}}\\\\7.2\ m/s = \sqrt{\frac{T}{2.9\ x\ 10^{-4}\ kg/m}}\\\\(51.84\ m^2/s^2)(2.9\ x\ 10^{-4}\ kg/m) = T

<u>T = 0.015 N</u>

<u></u>

(b)

The mass to be hung is:

T = Mg\\\\M = \frac{T}{g}\\\\M = \frac{0.015\ N}{9.8\ m/s^2}\\\\

<u>M = 1.53 x 10⁻³ kg = 1.53 g</u>

4 0
3 years ago
7. Two people are pushing a 40.0kg table across the floor. Person 1 pushes with a force of 490N
artcher [175]

Answer:

20.4 m/s^{2}

Explanation:

To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.

First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.

Let us determine the net force. Since the object is moving horizontally, we can ignore the weight and normal force, because they are vertical forces. The only horizontal forces we need to worry about are the applied force and force of friction.

Applied force = 1055 N (490 + 565)

Friction force= Unknown

To find the friction force, use the kinetic friction formula, Friction = μkN

μk is the coefficient, which the problem includes- it is 0.613.

N is the normal force, which we have to find.

*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.

Weight = mg = (40)(9.8) = 392 N

Normal force = 392 N

Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N

Friction = 240.296 N

Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N

Horizontal Net Force: 814.704 N

Now that we know the net force, plug in the numbers for the formula

∑F= ma.

814.704 = (40.0)(a)

*Divide on both sides)

a = 20.3676 m/s^2

Round it to 3 significant figures, to get:

20.4 m/s^{2}

7 0
3 years ago
A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h
Usimov [2.4K]

Explanation:

It is known that relation between torque and angular acceleration is as follows.

                    \tau = I \times \alpha

and,       I = \sum mr^{2}

So,      I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}

                       = 4 kg m^{2}

      \tau_{1} = 4 kg m^{2} \times \alpha_{1}

     \tau_{2} = I_{2} \alpha_{2}

So,      I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}

                     = 1 kg m^{2}

 as \tau_{2} = I_{2} \alpha_{2}

                   = 1 kg m^{2} \times \alpha_{2}        

Hence,     \tau_{1} = \tau_{2}

                  4 \alpha_{1} = \alpha_{2}

            \alpha_{1} = \frac{1}{4} \alpha_{2}

Thus, we can conclude that the new rotation is \frac{1}{4} times that of the first rotation rate.

8 0
3 years ago
What effect does gravity have on a person living on earth?
GREYUIT [131]
Pulling a person down so we stick to the surface
5 0
3 years ago
A. If we increase the wind velocity, the maximum vertical dispersal height and rate of diffusion will decrease____.
vovikov84 [41]

Answer:

a.If we increase the wind velocity, the maximum vertical dispersal height will decrease, while the rate of diffusion will increase

b.If we increase the humidity, the maximum vertical dispersal height will increase after 24 hours.

c.If we increase the lapse rate, the maximum vertical dispersal height of the pollutants will increase

Explanation:

a.If we increase the wind velocity, the maximum vertical dispersal height will decrease, while the rate of diffusion will increase

b.If we increase the humidity, the maximum vertical dispersal height will increase after 24 hours.

c.If we increase the lapse rate, the maximum vertical dispersal height of the pollutants will increase

8 0
3 years ago
Read 2 more answers
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