<h2>
Answer:</h2>
<em><u>Velocity of throwing arrow = 43.13 m/s.</u></em>
<h2>
Explanation:</h2>
In the question,
Let us say the height from which the arrow was shot = h
Distance traveled by the arrow in horizontal = 61 m
Angle made by the arrow with the ground = 2°
So,
From the <u>equations of the motion</u>,
Now,
Also,
Finally, the angle made is 2 degrees with the horizontal.
So,
Final horizontal velocity = v.cos20°
Final vertical velocity = v.sin20°
Now,
u = v.cos20° (No acceleration in horizontal)
Also,
So,
We can say that,
<em><u>Therefore, the velocity with which the arrow was shot by the archer is 43.13 m/s.</u></em>
Answer:
0.256 hours
Explanation:
<u>Vectors in the plane
</u>
We know Office A is walking at 5 mph directly south. Let 
be its distance. In t hours he has walked
Office B is walking at 6 mph directly west. In t hours his distance is
Since both directions are 90 degrees apart, the distance between them is the hypotenuse of a triangle which sides are the distances of each office
This distance is known to be 2 miles, so
t is approximately 15 minutes
Answer:
a= -1.2 m/s^2
Vi= 6.5 m/s
Vf= 0 m/s
t= 0-6.5/-1.2= <u>5.45 Sec</u>
Explanation:
Answer:
<em>765,000Joules or 765kJ</em>
Explanation:
The Quantity of heat required is expressed as;
Q = (mcΔt)al + (mcΔt)water
m is the mass
c is specific heat capacity
Δt is the change in temperature
Q = (3(900)(90-5)) + (1.5(4200)(90-5))
Q = 2700*85 + 6300*85
Q = (2700+6300)85
Q = 9000*85
<em>Q = 765,000</em>
<em>Hence the amount of energy needed is 765,000Joules or 765kJ</em>
Answer:
(a) 1.21 m/s² (b) 1.75 m/s²
Explanation:
The initial speed of the car, u = 17.8 m/s
Case 1.
Final speed of the car, v = 23.5 m/s
Time, t = 4.68-s
Acceleration = rate of change of velocity
Case 2.
Final speed of the car, v = 15.3 m/s
Hence, this is the required solution.
