Answer:
a)
Y0 = 0 m
Vy0 = 15 m/s
ay = -9.81 m/s^2
b) 7.71 m
c) 3.06 s
Explanation:
The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards
Y(0) = 0 m
Vy(0) = 15 m/s
ay = -9.81 m/s^2 (negative because it points down)
Since acceleration is constant we can use the equation for uniformly accelerated movement:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
To find the highest point we do the first time derivative (this is the speed:
V(t) = Vy0 + a * t
We equate this to zero
0 = Vy0 + a * t
0 = 15 - 9.81 * t
15 = 9.81 * t
t = 0.654 s
At this time it will have a height of:
Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m
The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.
0 = Y0 + Vy0 * t + 1/2 * a * t^2
0 = 0 + 15 * t - 1/2 * 9.81 t^2
0 = 15 * t - 4.9 * t^2
0 = t * (15 - 4.9 * t)
t1 = 0 This is the moment it jumped into the air
0 = 15 - 4.9 * t2
15 = 4.9 * t2
t2 = 3.06 s This is the moment when it falls again.
3.06 - 0 = 3.06 s
Answer:
t_{out} = 
t_{in}, t_{out} = 
Explanation:
This in a relative velocity exercise in one dimension,
let's start with the swimmer going downstream
its speed is
The subscripts are s for the swimmer, r for the river and g for the Earth
with the velocity constant we can use the relations of uniform motion
= D / 
D = v_{sg1} t_{out}
now let's analyze when the swimmer turns around and returns to the starting point
= D / 
D = v_{sg 2} t_{in}
with the distance is the same we can equalize
t_{out} = t_{in}
t_{out} = t_{in}
This must be the answer since the return time is known. If you want to delete this time
t_{in}= D / 
we substitute
t_{out} = \frac{v_s - v_r}{v_s+v_r} ()
t_{out} =
Answer:
travels along a curve due to uniform gravity
Explanation:
General relativity
The theory was developed by Albert Einstein , it is the theory of gravitation , according to which , the gravitation effect which is observed between the masses , is the resultant of their warping of the space time .
In general relativity , the equivalence principle is the equivalence of inertial gravitational mass .
According to which , the light bends due to gravitational forces ,therefore , the light travels along a curve because of uniform gravity .
Answer:
1,800kg
Explanation:
Force = Mass x Acceleration
F = m x a
15840N = m x 8.8
8.8 x m = 15840
m = 15840/8.8
= 1,800kg
Answer:
The highest electric field is experienced by a 2 C charge acted on by a 6 N electric force. Its magnitude is 3 N.
Explanation:
The formula for electric field is given as:
E = F/q
where,
E = Electric field
F = Electric Force
q = Charge Experiencing Force
Now, we apply this formula to all the cases given in question.
A) <u>A 2C charge acted on by a 4 N electric force</u>
F = 4 N
q = 2 C
Therefore,
E = 4 N/2 C = 2 N/C
B) <u>A 3 C charge acted on by a 5 N electric force</u>
F = 5 N
q = 3 C
Therefore,
E = 5 N/3 C = 1.67 N/C
C) <u>A 4 C charge acted on by a 6 N electric force</u>
F = 6 N
q = 4 C
Therefore,
E = 6 N/4 C = 1.5 N/C
D) <u>A 2 C charge acted on by a 6 N electric force</u>
F = 6 N
q = 2 C
Therefore,
E = 6 N/2 C = 3 N/C
E) <u>A 3 C charge acted on by a 3 N electric force</u>
F = 3 N
q = 3 C
Therefore,
E = 3 N/3 C = 1 N/C
F) <u>A 4 C charge acted on by a 2 N electric force</u>
F = 2 N
q = 4 C
Therefore,
E = 2 N/4 C = 0.5 N/C
The highest field is 3 N, which is found in part D.
<u>A 2 C charge acted on by a 6 N electric force</u>
