At 100 km/hr, the car's kinetic energy is
KE = (1/2) (mass) (speed)²
KE = (1/2) (1575 kg) ( [100 km/hr] x [1000 m/km] x [1 hr/3600 sec] )²
KE = (787.5 kg) (27.78 m/s)²
KE = 607,639 Joules
In order to deliver this energy in 2.9 seconds, the engine must supply
(607,639 J / 2.9 sec) = 209,531 watts
<em>Power = 281 HP</em>
Density = (mass) / (volume)
4,000 kg/m³ = (mass) / (0.09 m³)
Multiply each side
by 0.09 m³ : (4,000 kg/m³) x (0.09 m³) = mass
mass = 360 kg .
Force of gravity = (mass) x (acceleration of gravity)
= (360 kg) x (9.8 m/s²)
= (360 x 9.8) kg-m/s²
= 3,528 newtons .
That's the force of gravity on this block, and it doesn't matter
what else is around it. It could be in a box on the shelf or at
the bottom of a swimming pool . . . it's weight is 3,528 newtons
(about 793.7 pounds).
Now, it won't seem that heavy when it's in the water, because
there's another force acting on it in the upward direction, against
gravity. That's the buoyant force due to the displaced water.
The block is displacing 0.09 m³ of water. Water has 1,000 kg of
mass in a m³, so the block displaces 90 kg of water. The weight
of that water is (90) x (9.8) = 882 newtons (about 198.4 pounds),
and that force tries to hold the block up, against gravity.
So while it's in the water, the block seems to weigh
(3,528 - 882) = 2,646 newtons (about 595.2 pounds) .
But again ... it's not correct to call that the "force of gravity acting
on the block in water". The force of gravity doesn't change, but
there's another force, working against gravity, in the water.
Answer:
Yes
Explanation:
The farther something is from the center of mass of an object such as a planet, the lower the gravitational force between them
F = GMm/d²
B) 37 V
The potential difference between point a and point b is equal to 37 V
<h3>
Explanation:</h3>
The potential difference across a and b can be calculated by finding the equivalent resistance of the resistors R1, R2 and R3 and then by applying Ohm's law.
Given:
R1 = 3.0 Ω
R2 = 8.0 Ω
R3 = 10.0
Current = I = 5 A
<u>To calculate equivalent resistance of the circuit: </u>
R2 and R3 are connected in parallel configuration. The equivalent resistance of R2 and R3 can be calculated as
![\frac{1}{R_p} =\frac{1}{R_2} + \frac{1}{R_3} \\\\\frac{1}{R_p}=\frac{1}{8} + \frac{1}{10}= \frac{5+4}{40} = \frac{9}{40} \\\\ \frac{1}{R_p}=\frac{9}{40} \\\\R_p=\frac{40}{9}\ ohms](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BR_p%7D%20%3D%5Cfrac%7B1%7D%7BR_2%7D%20%2B%20%5Cfrac%7B1%7D%7BR_3%7D%20%5C%5C%5C%5C%5Cfrac%7B1%7D%7BR_p%7D%3D%5Cfrac%7B1%7D%7B8%7D%20%2B%20%5Cfrac%7B1%7D%7B10%7D%3D%20%5Cfrac%7B5%2B4%7D%7B40%7D%20%3D%20%5Cfrac%7B9%7D%7B40%7D%20%5C%5C%5C%5C%20%5Cfrac%7B1%7D%7BR_p%7D%3D%5Cfrac%7B9%7D%7B40%7D%20%5C%5C%5C%5CR_p%3D%5Cfrac%7B40%7D%7B9%7D%5C%20ohms)
Resistance R1 is in series with Rp. The equivalent series resistance is calculated as
![R_s = R1+R_p = 3+\frac{40}{9}= \frac{27+40}{9} = \frac{67}{9} = 7.44\ ohms](https://tex.z-dn.net/?f=R_s%20%3D%20R1%2BR_p%20%3D%203%2B%5Cfrac%7B40%7D%7B9%7D%3D%20%5Cfrac%7B27%2B40%7D%7B9%7D%20%3D%20%5Cfrac%7B67%7D%7B9%7D%20%3D%207.44%5C%20ohms)
Thus the given 3 resistor circuit can be written as an equivalent resistance Rs equal to 7.44 Ω through which current I passes.
By Ohm's law;
If V is the potential difference between a and b
Potential difference across points a and b = Current flowing through the circuit × Resistance between points a and b
![V = I\times R_s = 5\times 7.44 = 37.2\ V](https://tex.z-dn.net/?f=V%20%3D%20I%5Ctimes%20R_s%20%3D%205%5Ctimes%207.44%20%3D%2037.2%5C%20V)
37.2 V ≈ 37 V
Therefore, the potential difference between point a and point b is equal to 37 V