Answer:
89.11kg
Explanation:
Note an object weighs less when in a fluid and the weight of the volume of the fluid displaced is known as the upthrust.
Now, the person is going to displace the volume 89/1025 =0.087m3 { from density D = mass(M)/volume(V)}
The weight of the fluid displaced is the density of the fluid × volume of fluid displaced.
The weight of the fluid=0.087m3× 1kg/me = 0.087kg
Now the weight of the fluid displaced is referred to as the upthrust.
Now the real weight - the apparent weight = the upthrust.
Hence the apparent weight = real weight - upthrust
Apparent weight = 89.2-0.087 = 89.11kg
<span>The drag force acts opposite to the direction the cars' motion and the propelling force is in the direction of the motion. These are the two forces acting on the car and since they are both equal and opposite in direction, they cancel out each other. Thus, the car is able to move at a constant speed without changing its direction or momentum.</span>
Answer:
Explanation:
The equation fo potential energy is PE = mgh, where m is the mass of the ball, g is the pull of gravity (constant at 9.8), and h is the max height of the ball. What we do not have here is that height. We need to first solve for it using one-dimensional equations. What we have to know above all else, is that the final velocity of an object at its max height is always 0. That allows us to use the equation
where vf is the final velocity and v0 is the initial velocity. We will find out how long it takes for the object to reach that max height first and then use that time to find out what that max height is. Baby steps here...
0 = 21.5 + (-9.8)t and
-21.5 = -9.8t so
t = 2.19 seconds (Keep in mind that if I used the rules correctly for sig fig's, the answer you SHOULD get is not one shown, so I had to adjust the sig fig's and break the rules. But you know what they say about rules...)
Now we will use that time to find out the max height of the object in the equation
Δx = 
and filling in:
Δx = 
which simplifies down a bit to
Δx = 47.1 - 23.5 so
Δx = 23.6 meters.
Now we can plug that in to the PE equation to find the PE of the object:
PE = (.19)(9.8)(23.6) so
PE = 43.9 J
Answer:
I think the reflection of light off of a shiny surface is the answer... Hope this helps
Explanation:
Answer:
Explanation:
Force = q ( v x B)
- 5.6 x 10⁻⁹ (v x - 1.25 k )
- 3.4x 10⁻⁷i + 7.4 x 10⁻⁷j
Let v = ai+bj +ck
Force = - 5.6 x 10⁻⁹ [(ai+bj +ck) x - 1.25 k )]
= - 5.6 x 10⁻⁹ ( 1.25aj - 1.25bi )
= - 7 a j + 7 b i
( 7bi - 7aj ) x 10⁻⁹
Comparing with given force
7b x 10⁻⁹ b = - 3.4 x 10⁻⁷
b = - 48.57
- 7 a x 10⁻⁹ = 7.4 x 10⁻⁷
a = - 105.7
velocity
= -105.7 i - 48.57 j + ck
b ) Component along k can not be obtained .
c ) v . F = ( -105.7 i - 48.57 j + ck ) . −(3.40×10−7N) ˆı +(7.40×10−7N) ˆȷ
= 105.7 x 3.4 x 10⁻⁷ - 48.57 x 7.4 x 10⁻⁷
= 359.38 x 10⁻⁷ - 359.38 x 10⁻⁷
=0
angle between v and F = 90 degree
