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2 years ago

A 3,000-N force gives an object an acceleration of 15 m/s^2. The mass of the object is

2 answers:

8 0

The correct answer is m= 200 kg

Explanation: The solution for solving the mass kg the object using Newton’s Law of Accerelation is F= ma.

Bingel [31]2 years ago

7 0

<h2>Answer:</h2>

<em>Hello, </em>

<h3><u>QUESTION)</u></h3>

According to the second Newton's Law,

m = 3000/15
m = 200 kg

The mass of the object is therefore 200 kg.

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A car driving on the highway is going 72mph After 3 hours how many miles will the car have driven ?

Sergeu [11.5K]

Answer:

216 miles

Explanation:

you times 72 by three

7 0

2 years ago

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A candle is placed 30 cm in front of a convex mirror with a focal length of 20 cm, as shown in the diagram. What is the distance

lora16 [44]

I think the answer is 60 cm.

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3 years ago

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The first artificial satellite to orbit the Earth was Sputnik I, launched October 4, 1957. The mass of Sputnik I was 83.5 kg, an

9966 [12]

Answer:

$-4.941*10^8J.$

Explanation:

To solve this exercise it is necessary to take into account the concepts related to gravitational potential energy, as well as the concept of perigee and apogee of a celestial body.

By conservation of energy we know that,

$\Delta U = \Delta_{perogee}-\Delta_{Apogee}$

Where,

$U= \frac{-GmM_e}{r}$

Replacing

$\Delta U = \frac{-GmM_e}{r_p}- \frac{-GmM_e}{r_a}$

$\Delta U = GmM_e (\frac{1}{r_A}-\frac{1}{r_p})$

Our values are given by,

$m = 85.5Kg$

$M_e = 5.97*10^{24}Kg$

$r_A = 7330Km$

$r_p = 6610Km$

$G = 6.67*10^{-11}Nm^2/Kg^2$

Replacing at the equation,

$\Delta U = (6.67*10^{-11})(85.5)(5.97*10^{24}) (\frac{1}{7330}-\frac{1}{6610})$

$\Delta U = -4.941*10^8J$

Therefore the Energy necessary for Sputnik I as it moved from apogee to perigee was $-4.941*10^8J.$

4 0

2 years ago

While taking off from an aircraft carrier, a jet starting from rest accelerates uniformly to a final speed of 40. meters per sec

mariarad [96]

The acceleration of the jet is $11.4 m/s^2$

Explanation:

Since the jet motion is a uniformly accelerated motion (=constant acceleration), we can use the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

For the jet in this problem, we have

u = 0 (it starts from rest)

v = 40 m/s (final velocity)

s = 70 m (length of the runway)

Solving for a, we find the acceleration:

$a=\frac{v-u}{t}=\frac{40^2-0}{2(70)}=11.4 m/s^2$

Learn more about acceleration:

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5 0

2 years ago

ASK YOUR TEACHER A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 51.5-gram mass is attach

Simora [160]

Answer:

0.114 kg or 114 g

Explanation:

From the diagram attaches,

Taking the moment about the fulcrum,

sum of clockwise moment = sum of anticlockwise moment.

Wd = W'd'

Where W = weight of the mass, W' = weight of the meter rule, d = distance of the mass from the fulcrum, d' = distance of the meter rule.

make W' the subject of the equation

W' = Wd/d'................ Equation 1

Given: W = mg = 0.0515(9.8) = 0.5047 N, d = (39.2-16) = 23.2 cm, d' = (49.7-39.2) = 10.5 cm

Substitute these values into equation 1

W' = 0.5047(23.2)/10.5

W' = 1.115 N.

But,

m' = W'/g

m' = 1.115/9.8

m' = 0.114 kg

m' = 114 g

6 0

2 years ago