Answer:
I = 5[amp]
Explanation:
Electrical power is defined as the product of voltage by current.
where:
P = power = 1150 [W]
V = voltage = 230 [V]
I = current [amp]
Now replacing:
![1150=230*I\\I=1150/230\\I=5[amp]](https://tex.z-dn.net/?f=1150%3D230%2AI%5C%5CI%3D1150%2F230%5C%5CI%3D5%5Bamp%5D)
A 15 [amp] fuse must be used. Always the fuse must be larger than the operating current, to protect the equipment from very high currents. above 15 [amp]
OF COURSE !
The gravitational force between two objects doesn't ONLY depend on the product of their masses. It also depends on the distance between them.
I'm not even going to work out the numbers for my example. I'm just going to state without proof that at the top of the 2nd frame, the gravitational force between you and your bowling ball is greater than the gravitational force between you and the whole darn Andromeda galaxy ! My reasoning is based on the fact that your bowling ball is maybe 1 foot from your center of mass, whereas the Andromeda galaxy is more like 2.5 million light years from it. That right there is going to give your bowling ball a big advantage when it comes to gravity !
False- the potential energy is force*distance and force is mass*acceleration so if there’s more mass, there’s more force, so there’s more potential energy
Hope that helps :)
Answer:
114.86%
Explanation:
In both cases, there is a vertical force equal to the sprinter's weight:
Fy = mg
When running in a circle, there is an additional centripetal force:
Fx = mv²/r
The net force is found with Pythagorean theorem:
F² = Fx² + Fy²
F² = (mv²/r)² + (mg)²
F² = m² ((v²/r)² + g²)
F = m √((v²/r)² + g²)
Compared to just the vertical force:
F / Fy
m √((v²/r)² + g²) / mg
√((v²/r)² + g²) / g
Given v = 12 m/s, r = 26 m, and g = 9.8 m/s²:
√((12²/26)² + 9.8²) / 9.8
1.1486
The force is about 114.86% greater (round as needed).
Answer:
(a)
(b) It won't hit
(c) 110 m
Explanation:
(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1
(b) The velocity of the car before the driver begins braking is
The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is
We can use the following equation of motion to calculate how far the car has travel since braking to stop
Also the distance from start to where the driver starts braking is
So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb
(c) The distance from the limb to where the car stops is 550 - 440 = 110 m
