Why is it dangerous*btw.....And because you can easily crash into another vehicle also because it's just plain stupid to jump anyway ...
<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:
a) T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)
b) T = 295.37 K
Explanation:
Given;
Initial temperature of tea T1 = 31 C
Initial temperature of ice T2 = 0 C
Mass of tea m1 = 0.89 kg
Mass of ice m2 = 0.075kg
The heat capacity of both water and tea c = 4186 J/(kg⋅K)
the latent heat of fusion for water is Lf = 33.5 × 10^4 J/kg
And T = the final temperature of the mixture
Heat loss by tea = heat gained by ice
m1c∆T1 = m2c∆T2 + m2Lf
m1c(T1-T) = m2c(T-T2) + m2Lf
m1cT1 - m1cT = m2cT - m2cT2 + m2Lf
m1cT + m2cT = m1cT1 + m2cT2 - m2Lf
T(m1c + m2c) = m1cT1 + m2cT2 - m2Lf
T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)
Substituting the values;
T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)
T = (0.89×4186×31 + 0.075×4186×0 - 0.075×33.5 × 10^4)/(0.89×4186 + 0.075×4186)
T = 22.37 °C
T = 273 + 22.37 K
T = 295.37 K
Answer:
B v1= 12.5m/s
Explanation:
As the acceleration is constant and time taken is also same and distance covered is halfed so the speed will also be halfed.
An object in motion stays in motion, while an object at rest will stay at rest, otherwise known as inertia. So, a rolling ball will stay in motion if it's moving, whereas if it's being held in you hand and resting, it won't!
