Q1. The answer is 8.788 m/s
V2 = V1 + at
V1 - the initial velocity
V2 - the final velocity
a - the acceleration
t - the time
We have:
V1 = 4.7 m/s
a = 0.73 m/s²
t = 5.6 s
V2 = ?
V2 = 4.7 + 0.73 * 5.6
V2 = 4.7 + 4.088
V2 = 8.788 m/s
Q2. The answer is 9.22 s
V2 = V1 + at
V1 - the initial velocity
V2 - the final velocity
a - the acceleration
t - the time
We have:
V2 = 0 (because it reaches a complete stop)
V1 = 4.7 m/s
a = -0.51 m/s²
t = ?
0 = 4.7 + (-0.51)*t
0 = 4.7 - 0.51t
0.51t = 4.7
t = 4.7 / 0.51
t = 9.22 s
Explanation :
Takumi wears sunscreen and a hat each time he works in the yard. This is to protect himself with the strong radiation coming from the sun. UV rays that are coming from the sun are the main cause of skin cancer.
Stochastic effects are the effects that are caused by chance. Cancer is one of the main stochastic effects.
So, the correct option is (b) "the severity of stochastic effects, such as cancer".
Answer:
Then the difference of weight between the two cars are:
Δw = 14210 - 5292 = 8918 N
Explanation:
An object's weigh due to the gravitational attraction force of the earth is:
w = mg
Where: m is the object's mass
g is the gravitational acceleration in the surface earth
g = 9.8 m/s2
The the ultralight car's weight is:


And the Honda Accord's weight is:


Then the difference of weight between the two cars are:
Δw = 14210 - 5292 = 8918 N
The solution to the questions are given as


- the direction of induced current will be Counterclock vise.
<h3>What is the direction of the
current induced in the loop, as viewed from above the loop.?</h3>
Given, $B(t)=(1.4 T) e^{-0.057 t}$




(b) 

c)
In conclusion, the direction of the induced current will be Counterclockwise.
Read more about current
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The combined momentum is 4000 kg m/s south
Explanation:
The total combined momentum of the two cars is given by the vector addition of the momenta of the two cars.
For this problem, we choose north as positive direction and south as negative direction.
The momentum of the first car travelling north is given by:

where
is the mass of the car
is its velocity
Substituting,

The momentum of the second car travelling south is given by:

where
is the mass of the car
is its velocity (negative because the car travels south)
Substituting,

And therefore, the combined momentum is

where the negative sign means the direction of the total momentum is south.
Learn more about momentum:
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