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kykrilka [37]
3 years ago
12

What would be the mass of an object that has a force of 88N and an acceleration of 12 m/s2?

Physics
1 answer:
MA_775_DIABLO [31]3 years ago
6 0

Answer:

7.33 kg

Explanation:

F = ma

88 N = m(12 m/s^2)

m = 7.33 kg

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A truck with 0.410 m radius tires travels at 25.0 m/s. What is the angular velocity of the rotating tires in radians per second?
Eduardwww [97]

relation between linear velocity and angular velocity is given as

v = R\omega

here

v = linear speed

R = radius

\omega = angular speed

now plug in all data in the equation

25.0 = 0.410 \omega

\omega  = \frac{25}{0.410}

\omega = 60.9 rad/s

so rotating speed is 60.9 rad/s

6 0
3 years ago
Brainly malfunction?|<br><br> I'm not sure why but...
Brut [27]

Answer:

its probably still trying to load ur next rank or whatever it did it to me too

Explanation:

6 0
3 years ago
Explain how grality and electric charge are different
stich3 [128]
A difference between the two forces is the fact that gravitation only attracts, while electrical forces attract when the electrical charges are opposite and repel if the charges are similar. Thus, gravitation is considered a monopole force, while electrostatics is a dipole force. Please mark me as a brainiest!!!
6 0
3 years ago
The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of
adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = ma_a (0.35)

1.8v^2(1.1) - 1200(9.8)(1.25) = 1200a(0.35)

1.8v^2(1.1) - 14700 = 420 a   ------- equation (1)

F_x = ma_x \\ \\ = 1.8v^2 = 1200 \ a             --------- equation (2)

Replacing equation 2 into equation 1 ; we have :

{1.1 * 1200 \ a} - 14700 = 420 a

1320 a - 14700 = 420 a

1320 a -  420 a =14700

900 a = 14700

a = 14700/900

a = 16.33 m/s²

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

5 0
3 years ago
a cart's initial velocity is +3.0 meters per second. What is its final velocity after accelerating at a rate of 1.5m/s2 for 8.0
Andreas93 [3]

The final velocity is +15.0 m/s

Explanation:

The motion of the cart is a uniformly accelerated motion (=at constant acceleration), therefore we can use the following suvat equation:

v=u+at

where

v is the velocity at time t

u is the initial velocity

a is the acceleration

t is the time

For the cart in this problem, we have:

u = +3.0 m/s (initial velocity)

a=1.5 m/s^2 (acceleration)

t = 8.0 s (time)

Substituting, we find the final velocity:

v=3.0+(1.5)(8.0)=+15.0 m/s

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

4 0
3 years ago
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