3 years ago

What two factors affect the rate of acceleration in an object

Physics

2 answers:

Leto [7]3 years ago

8 0

Newton's second law states that F=ma so that means force =mass×acceleration force and mass

jonny [76]3 years ago

4 0

Force and mass affect acceleration
F=ma

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Tema [17]

Answer:

hwjsjsjsjksksnsjkdkdnnd

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2 years ago

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If I could lift up to ten tons and I threw a ball the size of an orange but weighed a ton, to the ground, how big of an impact w

Anarel [89]

Answer:

The impact force is 98000 N.

Explanation:

mass = 10 tons

The impact force is the weight of the object.

Weight =mass x gravity

W = 10 x 1000 x 9.8

W = 98000 N

The impact force is 98000 N.

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2 years ago

Planet Nine is speculated to be on average 20 times farther away from the Sun than Neptune (on average distance from the Sun). H

saveliy_v [14]

Answer:

The distance is 55.636 billion miles, or 528.2 AU.

Explanation:

Since the distance from the Sun to Neptune is 2.7818 billion miles, the distance from the Sun to Planet Nine would be 20 times that, which is:

$d=(20)(2781800000\ miles)=55636000000\ miles$

or 55.636 billion miles.

Since 1 astronomical unit (AU) is 93 million miles, that distance is also:

$d=(55636000000\ miles)(\frac{1AU}{93000000\ miles})=598.2\ AU$

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3 years ago

A 25 kg mass is accelerated by a force at a rate of 5 m/s2. What is the magnitude of the force that accelerates the man?

svlad2 [7]

Force=mass*acceleration
F=ma
F=25*5
F=100 N

8 0

2 years ago

A 6 kg box with initial speed 5 m/s slides across the floor and comes to a stop after 1.9 s. What is the coefficient of kinetic

Ilia_Sergeevich [38]

Answer:

$\mu_k=0.27$

Explanation:

According to the free body diagram, in this case, we have:

$\sum F_x:-F_k=ma\\\sum F_y:N=mg$

Recall that the force of friction is given by:

$F_k=\mu_k N$

Replacing and solving for the coefficient of kinetic friction:

$-\mu_kN=ma\\-\mu_k(mg)=ma\\\mu_k=-\frac{a}{g}$

We have an uniformly accelerated motion. Thus, the acceleration is defined as:

$a=\frac{v_f-v_0}{t}\\a=\frac{0-5\frac{m}{s}}{1.9s}\\a=-2.63\frac{m}{s^2}$

Finally, we calculate $\mu_k$ :

$\mu_k=-\frac{-2.63\frac{m}{s^2}}{9.8\frac{m}{s^2}}\\\mu_k=0.27$

4 0

3 years ago