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3 years ago

What is the acceleration of a 349 kg object that moved with a force of 750 N?

Physics

1 answer:

zavuch27 [327]3 years ago

4 0

Answer:

<h3>The answer is 2.15 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

where

f is the force

m is the mass

From the question we have

$a = \frac{750}{349} \\ = 2.14899713...$

We have the final answer as

Hope this helps you

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A car takes off from rest takes of from rest and covers a distance of 80m on a straight road in 10s.Calculate the magnitude of i

hodyreva [135]

Initial velocity (u) = 0 m/s [the car was at rest]
Distance (s) = 80 m
Time (t) = 10 s
Let the magnitude of acceleration be a.
By using the equation of motion, $s = ut + \frac{1}{2} a {t}^{2}$ we get, $80 = 0 \times 10 + \frac{1}{2} \times a \times {10}^{2} \\ = > 80 = \frac{1}{2} \times 100a \\ = > 80 = 50a \\ = > a = \frac{80}{50} \\ = > a = 1.6$

Answer:

The magnitude of its acceleration is 1.6 m/s^2.

Hope you could get an idea from here.

Doubt clarification - use comment section.

4 0

3 years ago

Suppose you have a 75-watt light bulb that you leave turned on for two minutes. How much energy

victus00 [196]

Answer:

9000 J

Explanation:

Convert minutes to seconds.

2 min = 120 s

Power = energy / time

75 W = E / 120 s

E = 9000 J

5 0

2 years ago

Two boxes are at rest on a smooth, horizontal surface. The boxes are in contact with one another. If box 1 is pushed with a forc

Maslowich

Answer:

mass of box 1 = 2.20 kg

mass of box 2 = 5.93 kg

Explanation:

Let the mass of box 1 and box 2 is respectively

$m_1$ and $m_2$

so we will have

Force applied on box 1 then acceleration

$a = \frac{F}{m_1 + m_2}$

$1.50 = \frac{12.2}{m_1 + m_2}$

$m_1 + m_2 = 8.13$

Now we know that contact force between them in above case is given as

$F_n = m_2a$

$8.90 = m_2(1.5)$

$m_2 = 5.93 kg$

now we have

$m_1 = 2.20 kg$

8 0

3 years ago

One uniform ladder of mass 30 kg and 10 m long rests against a frictionless vertical wall and makes an angle of 60o with the flo

yuradex [85]

Answer:

μ = 0.37

Explanation:

For this exercise we must use the translational and rotational equilibrium equations.

We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive

let's write the rotational equilibrium

W₁ x/2 + W₂ x₂ - fr y = 0

where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances

cos 60 = x / L

where L is the length of the ladder

x = L cos 60

sin 60 = y / L

y = L sin60

the horizontal distance of man is

cos 60 = x2 / 7.0

x2 = 7 cos 60

we substitute

m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0

fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60

let's calculate

fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)

fr = (735 + 2450) / 8.66

fr = 367.78 N

the friction force has the expression

fr = μ N

write the translational equilibrium equation

N - W₁ -W₂ = 0

N = m₁ g + W₂

N = 30 9.8 + 700

N = 994 N

we clear the friction force from the eucacion

μ = fr / N

μ = 367.78 / 994

μ = 0.37

3 0

3 years ago

Check all that are true about waterfalls

Lerok [7]

Waterfalls are created when a river flows following a descending rapid slope. The waterfall, then, flows from the source (where it starts) to the mouth (where it ends).
Waterfalls are created when the erosion of the rocks at the bottom of the slope is more powerful than the erosion of the rocks on the top.
After many years the water is able to erode the rocks on the top as well, and the waterfall slowly disappears.

Therefore the options that apply are:
b) waterfalls move towards their mouth;
c) the top or cap rock is resistant to erosion;
f) waterfalls indicate a youthful river

4 0

4 years ago