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2 years ago

First law of equilibrium

Physics

1 answer:

valentina_108 [34]2 years ago

5 0

Answer:

For an object to be an equilibrium it must be experiencing no acceleration.

Explanation:

Hope it helps.

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A projectile is launched with an initial speed of 60.0 m/s at an angle of 30.0° above the horizontal. The projectile lands on a

vodka [1.7K]

Answer:

51.96 m/s^-1

Explanation:

a) see the attachment

b) As we know the velocity of the projectile has two component, horizontal velocity v_ox. and vertical velocity v_oy as shown in the figure. At the highest point of the trajectory, the projectile has only horizontal velocity and vertical velocity is zero. Therefore at the highest point of the trajectory, the velocity of the projectile will be

v_ox=v_o*cosФ

=60*cos (30)

= 51.96 m/s^-1

3 0

3 years ago

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Do you want robot referees in your favorite class?

jeka94

Answer:

Sounds cool.. but what do they do?

Explanation:

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3 years ago

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If the distance d (in meters) traveled by an object in time t (in seconds) is given by the formula d = A + Bt^2, the SI units of

Yuliya22 [10]

Answer:

The SI units of the “A” is m (meters)

The SI units of the “B” is m/s^2

Explanation:

Given the distance = d meters.

Time taken to travel = t (seconds)

Function of the distance, d = A + Bt^2

Now we have given the above information and from the given distance function, we have to find the SI units of the A and B. Here, below are the SI units.

Thus, the SI units of the “A” is = m (meters)

The SI units of the “B” is = m/s^2

6 0

3 years ago

Brutus, a champion weight lifter, raises 220 kg a distance of 3.10 m. (a) how much work is done by brutus lifting the weights? j

Nana76 [90]

A) work = force * distance
mass is not a force, weight is, so we have to find the weight of the block.
Weight = mg
Weight = (220kg)(9.8)
Weight = 2156N
Work = 2156N * 3.10m
work = 6683.6J
b) Since he is holding the weights, it's not moving, therefore, he doesn't do any work
c) The answer is still the same amount of work when he lifted them.
d) The answer is no since when he let go the weight, he doesn't apply any force to the weight.
e) P = work/time
P = 6683.6J / 2.1s
P = 3182.67 watts

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2 years ago

Desperate for help! please!!!!!!!!

lys-0071 [83]

$0.4029 \mathrm{m} / \mathrm{s}^{2}$ is the acceleration of the box.

<u>Explanation:</u>

Given data:

Mass of the box = 3.74 kg

Flat friction-less ground is pulled forward by a 4.20 N force at a 50.0 degree angle and pulled back by a 2.25 N force at a 122 degree angle.

First, we need to find the net horizontal force acting on the box. With the given data, the equation can be formed as below. Net horizontal force acting on the box (F) is given by

$F=\left(4.20 \times \cos 50^{\circ}\right)+\left(2.25 \times \cos 122^{\circ}\right)$

$F=(4.20 \times 0.64278)+(2.25 \times-0.5299)$

F = 2.699676 – 1.192275 = 1.507 N

Next, find acceleration of the box using Newton's second law of motion. This states that the link between mass (m) of an objects and the force (F) required to accelerate it. The equation can be given as

$F=m \times acceleration\ (a)$

$\text {acceleration }(a)=\frac{F}{m}=\frac{1.507}{3.74}=0.4029 \mathrm{m} / \mathrm{s}^{2}$

8 0

3 years ago