Answer
The force on the left across the lab table.
Explanation
The Newton's third law of motion states that; <em>action and reaction are equal but a ct in opposite direction. </em>
When the block of is pulled on the right with a force of X Newtons then there is a force -X Newtons applying and equal force on the left. For every action there must be a reaction with is equal and applying in the opposite direction.
So, if the block is pulled on the right by a force of 8 N there is another equal force applying on the left.
Answer:
a = 2 m/s2
Explanation:
we know from newtons 2nd law
F = ma.
we also know that from hookes law we have
F = kx
equate both value of force to get value of acceleration
kx = ma,
where,
k is spring constant = 8.0 N/m
x is maximum displacement 0.10 m
m is mass of object 0.40 kg
a = \frac{kx}{m}
= \frac{8 *0 .10}{0.40}
a = 2 m/s2
Answer:
Explanation:
First, It's important to remember F = ma, and in this problem m = 13.3 kg
This can be reduced to a simple system of equations problem. Now if they are both going the same way then we add them, while if they are going the opposite way we subtract them. So let's call them F1 and F2, with F1 arger than F2. Now, When we add them together F1+F2 = (.723 m/s^2)*13.3kg and then when we subtract them, and have the larger one pushing toward the east, let's call F1 the larger one, F1-F2 = (.493 m/s^2)*13.3kg.
Can you solve this system of equations seeing them like this, or do you need more help?
Answer:
98.13m
Explanation:
Complete question
Daniel is 50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the building
CHECK THE ATTACHMENT
From the figure, using trigonometry
Tan(θ ) = opposite/adjacent
Where Angle (θ )= 63°
Opposite= X = height of the building
Adjacent= 50 m
Then substitute the values we have
Tan(63)= X/50
1.9626= X/50
X= 1.9626 × 50
X= 98.13m
Hence, the height of the building is 98.13m
Answer:
It's the 3rd option
Explanation:
Wind is caused by the differences in air pressure on Earth's surface.
