The acceleration that the net force will cause is 7.5 m/s²
To solve this problem, first, we need to find the net force.
⇒ Equation:
F = ma................ Equation 1
⇒ Where:
- m = mass of the box
- a = acceleration of the box
From the question,
⇒ Given;
⇒ Substitute these values into equation 1
We also use the same equation 1 to find the acceleration that the net force will cause to a box of mass 2.8 kg.
⇒ make a the subject in equation 1
- a = F/m................. Equation 2
⇒ Substitute the appropriate values into equation 2
Hence, The acceleration that the net force will cause is 7.5 m/s²
Learn more about acceleration here: brainly.com/question/605631
Answer:
I didn't quite understand your question, but the moon jelly population will increase if there are either more zooplankton or less sea turtles.
More zooplankton means more food => higher population
Less sea turtles means that the moon jelly will be lest often consumed => higher population
Answer:
The acceleration of the crate is 
.
Explanation:
Given that,
Force, F = 750 N
Mass of the crate, m = 250 kg
The coefficient of friction is 0.12.
We need to find the acceleration of the crate. The net force acting on the crate is given by :
f is frictional force, 
So, the acceleration of the crate is . Hence, this is the required solution.
Answer:
a. 8.136 × 10¹³ km b. 2.669 × 10¹⁹ feet
Explanation:
a. What is this distance in kilometers?
Since Sirius A is 8.6 light years away from Earth and one light year = distance travelled by light in a year = 3 × 10⁸ m/ s × 365 days/year × 24 hr/day × 3600 s/hr = 9.4608 × 10¹⁵ m = 9.4608 × 10¹² km.
Then 8.6 light years = 8.6 × 1 light year
= 8.6 × 9.4608 × 10¹² km
= 81.363 × 10¹² km
= 8.1363 × 10¹³ km
≅ 8.136 × 10¹³ km
b. What is this distance in feet?
Since 1 meter = 3.28 feet,
8.6 light years = 8.1363 × 10¹² km
= 8136.3 × 10¹⁵ m
= 8136.3 × 10¹⁵ × 1 m
= 8136.3 × 10¹⁵ × 3.28 feet
= 26687.1 × 10¹⁵ feet
= 2.66871 × 10¹⁹ feet
≅ 2.669 × 10¹⁹ feet
Answer:
m1 = 20g (= 0.02 kg)
Mass of pistol, m2 = 2 kg
Initial velocity of the bullet (u1) and pistol (u2) = 0
Final velocity of the bullet, v1 = +150m s-1
Let v be the recoil velocity of the pistol.
Total momentum of the pistol and bullet after it is fired is
= (0.02 kg x 150 m s-1) + (2 kg x v m s-1)
= (3 + 2v) kg m s-1
Total momentum after the fire = Total momentum before the fire
3 + 2v = 0
→v = -1.5 m/s
