All categories

3 years ago

Can someone help with number 4 please

Physics

2 answers:

Neko [114]3 years ago

6 0

Here we have -
Acceleration (a) = ?
Initial velocity (u) = 0 m/s (runner starts from rest)
Distance (s) = 12 m
Time (t) = 4 seconds

So, using the second equation of motion, we have-

$s = ut + \frac{1}{2}at^{2}$

$12 = 0*4 + \frac{1}{2}*a*4 ^{2}$

$12 = 0+8a$

$\frac{12}{8} = a$

$1.5m/ s^{2}$

So, option (2) is correct.

kondaur [170]3 years ago

4 0

The way I look at these things is like this:

-- The runner covered 12 meters in 4 seconds.
Average speed = 3 meters per second.

-- Speed at the beginning = zero.
In order to make the average 3, Speed at the end = 6 meters per second.

-- Speed increased from zero to 6 meters per second in 4 seconds.
It must have increased 1.5 meters per second each second.

That's choice-#2.

You might be interested in

A luggage handler pulls a suitcase of mass 19.6 kg up a ramp inclined at an angle 24.0 ∘ above the horizontal by a force F⃗ of m

Dvinal [7]

(a) 638.4 J

The work done by a force is given by

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and the displacement

Here we want to calculate the work done by the force F, of magnitude

F = 152 N

The displacement of the suitcase is

d = 4.20 m along the ramp

And the force is parallel to the displacement, so $\theta=0^{\circ}$ . Therefore, the work done by this force is

$W_F=(152)(4.2)(cos 0)=638.4 J$

b) -328.2 J

The magnitude of the gravitational force is

W = mg

where

m = 19.6 kg is the mass of the suitcase

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting,

$W=(19.6)(9.8)=192.1 N$

Again, the displacement is

d = 4.20 m

The gravitational force acts vertically downward, so the angle between the displacement and the force is

$\theta= 90^{\circ} - \alpha = 90+24=114^{\circ}$

Where $\alpha = 24^{\circ}$ is the angle between the incline and the horizontal.

Therefore, the work done by gravity is

$W_g=(192.1)(4.20)(cos 114^{\circ})=-328.2 J$

c) 0

The magnitude of the normal force is equal to the component of the weight perpendicular to the ramp, therefore:

$R=mg cos \alpha$

And substituting

m = 19.6 kg

g = 9.8 m/s^2

$\alpha=24^{\circ}$

We find

$R=(19.6)(9.8)(cos 24)=175.5 N$

Now: the angle between the direction of the normal force and the displacement of the suitcase is 90 degrees:

$\theta=90^{\circ}$

Therefore, the work done by the normal force is

$W_R=R d cos \theta =(175.4)(4.20)(cos 90)=0$

d) -194.5 J

The magnitude of the force of friction is

$F_f = \mu R$

where

$\mu = 0.264$ is the coefficient of kinetic friction

R = 175.5 N is the normal force

Substituting,

$F_f = (0.264)(175.5)=46.3 N$

The displacement is still

d = 4.20 m

And the friction force points down along the slope, so the angle between the friction and the displacement is

$\theta=180^{\circ}$

Therefore, the work done by friction is

$W_f = F_f d cos \theta =(46.3)(4.20)(cos 180)=-194.5 J$

e) 115.7 J

The total work done on the suitcase is simply equal to the sum of the work done by each force,therefore:

$W=W_F + W_g + W_R +W_f = 638.4 +(-328.2)+0+(-194.5)=115.7 J$

f) 3.3 m/s

First of all, we have to find the work done by each force on the suitcase while it has travelled a distance of

d = 3.80 m

Using the same procedure as in part a-d, we find:

$W_F=(152)(3.80)(cos 0)=577.6 J$

$W_g=(192.1)(3.80)(cos 114^{\circ})=-296.9 J$

$W_R=(175.4)(3.80)(cos 90)=0$

$W_f =(46.3)(3.80)(cos 180)=-175.9 J$

So the total work done is

$W=577.6+(-296.9)+0+(-175.9)=104.8 J$

Now we can use the work-energy theorem to find the final speed of the suitcase: in fact, the total work done is equal to the gain in kinetic energy of the suitcase, therefore

$W=\Delta K = K_f - K_i\\W=\frac{1}{2}mv^2\\v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(104.8)}{19.6}}=3.3 m/s$

6 0

3 years ago

A college student is working on her physics homework in her dorm room. her room contains a total of 6.0×1026 gas molecules. as s

ella [17]

6.6 degrees C Let's model the student as a 125 w furnace that's been operating for 11 minutes. So 125 w * 11 min = 125 kg*m^2/s^3 * 11 min * 60 s/min = 82500 kg*m^2/s^2 = 82500 Joule So the average kinetic energy increase of each gas molecule is 82500 J / 6.0x10^26 = 1.38x10^-22 J Now the equation that relates kinetic energy to temperature is: E = (3/2)Kb*Tk E = average kinetic energy of the gas particles Kb = Boltzmann constant (1.3806504Ă—10^-23 J/K) Tk = Kinetic temperature in Kelvins Notice the the energy level of the gas particles is linear with respect to temperature. So we don't care what the original temperature is, we just need to know by how much the average energy of the gas particles has increased by. So let's substitute the known values and solve for Tk E = (3/2)Kb*Tk 1.38x10^-22 J = (3/2)1.3806504Ă—10^-23 J/K * Tk 1.38x10^-22 J = 2.0709756x10^-23 J/K * Tk 6.64 K = Tk Rounding to 2 significant digits gives 6.6K. So the temperature in the room will increase by 6.6 degrees K or 6.6 degrees C, or 11.9 degrees F.

7 0

3 years ago

Define faith in your own words.

krok68 [10]

Answer:

Faith, to me is almost like hope but for faith you need to believe. Some people associate the word faith with religion. But that's not it. Faith is really just a stronger term than believing someone. Like you can believe in someone but not have faith in that same person. You can have faith in someone but to do that you have to believe and trust them.

Explanation:

6 0

2 years ago

Shinra wants to find and save his younger brother Sho. In order to do this

Debora [2.8K]

Answer:

My answer is 250m/sec

Explanation:

the formula is clearly

Speed = Distance

Time

Speed = 500meters (Divide the two)

2seconds

Speed = 250m/sec

4 0

3 years ago

Read 2 more answers

Law of motion that says for every action there is an equal and opposite

hodyreva [135]

Answer: Newton's third law

Formally stated, Newton's third law is: For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object.

Explanation:

4 0

3 years ago

Read 2 more answers