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4 years ago

Which object has a net force of -68

2 answers:

3 0

The net force is the vector sum of all the forces that act upon an object. That is to say, the net force is the sum of all the forces, taking into account the fact that a force is a vector and two forces of equal magnitude and opposite direction will cancel each other out.

Paladinen [302]4 years ago

3 0

Answer:

The correct answer is C. on e2020.

Explanation:

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An elevator filled with passengers has a mass of 1.70×103kg. (a) The elevator accelerates upward from rest at a rate of 1.20m/s2

Anton [14]

Answer with Explanation:

We are given that mass of an elevator filled with passengers = $1.70\times 10^3 Kg$

a.a= $1.2 m/s^2$

t=1.50 s

Initial velocity=0

We have to calculate the tension in the cable supporting the elevator.

According to newton's second law,the net force is given by

$F_net=\sum F=ma$

$F-mg=ma$

Substitute the values then we get

$F=ma+mg=m(a+g)=1.7\times 10^3(1.2+9.8)=1.87\times 10^4 N$

Hence, the tension in the cable supporting the elevator= $1.87\times 10^4 N$

b.t=8.5 s

We have to find the tension in the table during this time.

Fnet=0

F=W= $1.7\times 10^3\times 9.8=1.67\times 10^4 N$

Hence, the tension in the cable during this time = $1.67\times 10^4 N$

c. Deceleration= $0.6 m/s^2$

t=3 s

We have to find the tension in the cable during deceleration.

$F-mg=-ma$

$F=mg-ma=m(g-a)$

$F=1.7\times 10^3(9.8-0.6)=1.56\times 10^4 N$

Hence, the tension in the cable during deceleration= $1.56\times 10^4 N$ .

d.We have to find that the height of elevator moved above its original staring point .

We have to find the final velocity.

$y_1=ut_1+\frac{1}{2}a_1t_1^2$

$y_1=0(1.50+\frac{1}{2}(1.2)(1.5)=1.35 m$

Final velocity

$v_1=a_1t_1$

$v_1=(1.2)(1.5)=1.8 m/s$

$y_2=v_1t_1=(1.8)(8.5)=15.3 m$

The third distance $y_3$ is up in the direction because the elevator decelerate upward which mean the distance is positive and is given by

$y_3=(1.8)(3)-\frac{1}{2}(0.6)(3)^2=2.7 m$

The total distance moved by elevator from its original position

$d=1.35+1.8+2.7=19.35 m$

Final velocity=0

5 0

4 years ago

Monochromatic light from a distant source is incident on a slit 0.705 mm wide. On a screen 2.13 m away, the distance from the ce

vampirchik [111]

Answer:

492.183 nm

Explanation:

x = Distance from the central maximum to the first minimum = 1.35 mm

l = Distance of screen = 2.13 m

d = Distance of gap = 0.705 mm

m = Order = 1

We have the relation

$tan\theta=\dfrac{x}{l}\\\Rightarrow \theta=tan^{-1}\dfrac{1.35\times 10^{-3}}{2.13}\\\Rightarrow \theta=0.04^{\circ}$

$dsin\theta=m\lambda\\\Rightarrow \lambda=\dfrac{dsin\theta}{m}\\\Rightarrow \lambda=\dfrac{0.705\times 10^{-3}\times sin0.04}{1}\\\Rightarrow \lambda=4.92183\times 10^{-7}=492.183\ nm$

The wavelength of the light is 492.185 nm

5 0

3 years ago

Read 2 more answers

Which of the following is an example of acceleration?

Naddika [18.5K]

Answer:

Explanation:

When a satellite is orbiting the earth , a constant force is being applied on it which means it must has acceleration. Also the direction of satellite is always being changed when it is orbitting to there is always change in the velocity vector which means acceleration.

You can view in the attached diagram to understand how the velocity is being changed.

7 0

3 years ago

Why must the difference between the water level in the eudiometer tube and the water level in the beaker be measured

Bas_tet [7]

Answer:

Explanation:

The difference between the water level in the eudiometer tube and the water level in the beaker must be measured because we have to put into consideration, the pressure of the gases in the eudiometer tube. This said pressure of gas in the eudiometer must equal the atmospheric pressure. If or by chance, the water levels happens not to be at the same height, then this is not the case. And then, as a result, in order to account for the difference between both, while also being able to get accurate results, you have to find the difference or subtract the water levels and then go ahead in converting them to mmHg.

3 0

3 years ago

This problem has been solved!

RideAnS [48]

Answer:

Option (a)

Explanation:

We will discard options that don't fit the situation:

Option b: Incorrect since if the driver "hits the gas" then velocity is augmenting and it's not constant.

Option c and d: Incorrect since the situation doesn't give us any information that could be related directly to the terrain or movement direction.

Option a: Correct. At stage 1 we can assume the driver was going at constant speed which means acceleration is constantly zero. At stage 2 we can assume the driver augmented speed linearly, this is, with constant positive acceleration. At stage 3 we can assume the driver slowed the speed linearly, with constant negative acceleration.

6 0

3 years ago