Protons have a positive charge which is indicated by a + sign.
Hence, the answer is C.
- Let, the maximum height covered by projectile be
![\sf{H_m}](https://tex.z-dn.net/?f=%5Csf%7BH_m%7D)
![\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }}](https://tex.z-dn.net/?f=%5Cpurple%7B%20%5Clongrightarrow%20%20%5Cbf%7Bh_m%20%3D%20%20%5Cdfrac%7B%20%7Bv%7D%5E%7B2%7D%20%5C%3A%20%7Bsin%7D%5E%7B2%7D%20%5Ctheta%20%20%7D%7B2g%7D%20%7D%7D%20)
- Projectile is thrown with a velocity = v
- Angle of projection = θ
- Velocity of projectile at a height half of the maximum height covered be
![\sf{v_0}](https://tex.z-dn.net/?f=%20%5Csf%7Bv_0%7D)
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Then –
![\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}](https://tex.z-dn.net/?f=%5Cpink%7B%20%20%5Clongrightarrow%20%5Cbf%7B%20%5Cdfrac%7Bh_m%7D%7B2%7D%20%20%3D%20%5Cdfrac%7B%20%7Bv_0%7D%5E%7B2%7D%20%5C%3A%20%7Bsin%7D%5E%7B2%7D%20%5Ctheta%20%20%7D%7B2g%7D%20%7D%7D)
![\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }](https://tex.z-dn.net/?f=%20%20%5Clongrightarrow%20%5Csf%7B%20%5Cdfrac%7B%20%7Bv%7D%5E%7B2%7D%20%20%5C%3A%20%7Bsin%7D%5E%7B2%7D%20%5Ctheta%20%20%7D%7B2g%7D%20%5Ctimes%20%20%5Cdfrac%7B1%7D%7B2%7D%20%20%3D%20%20%5Cdfrac%7B%20%7Bv_0%7D%5E%7B2%7D%20%5C%3A%20%7Bsin%7D%5E%7B2%7D%20%5Ctheta%20%20%7D%7B2g%7D%20%7D)
![\qquad](https://tex.z-dn.net/?f=%5Cqquad)
![\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }](https://tex.z-dn.net/?f=%5Clongrightarrow%20%20%5Csf%7B%20%5Cdfrac%7B%20%7Bv%7D%5E%7B2%7D%20%20%5C%3A%20%7Bsin%7D%5E%7B2%7D%20%5Ctheta%20%20%7D%7B4g%7D%20%20%3D%20%20%5Cdfrac%7B%20%7Bv_0%7D%5E%7B2%7D%20%5C%3A%20%7Bsin%7D%5E%7B2%7D%20%5Ctheta%20%20%7D%7B2g%7D%20%7D)
![\qquad](https://tex.z-dn.net/?f=%5Cqquad)
![\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }](https://tex.z-dn.net/?f=%5Clongrightarrow%20%20%5Csf%7B%20%5Cdfrac%7B%20%7Bv%7D%5E%7B2%7D%20%20%5C%3A%20%7Bsin%7D%5E%7B2%7D%20%5Ctheta%20%20%7D%7B2%7D%20%20%3D%20%20%20%7Bv_0%7D%5E%7B2%7D%20%5C%3A%20%7Bsin%7D%5E%7B2%7D%20%5Ctheta%20%7D)
![\qquad](https://tex.z-dn.net/?f=%5Cqquad)
![\longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }](https://tex.z-dn.net/?f=%20%5Clongrightarrow%20%20%5Csf%7B%20%5Cdfrac%7B%20%7Bv%7D%5E%7B2%7D%20%7D%7B2%7D%20%20%3D%20%20%20%7Bv_0%7D%5E%7B2%7D%20%7D)
![\qquad](https://tex.z-dn.net/?f=%5Cqquad)
![\longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }](https://tex.z-dn.net/?f=%20%20%5Clongrightarrow%20%5Cbf%7Bv_0%20%3D%20%20%20%5Csqrt%7B%20%5Cdfrac%7B%20%7Bv%7D%5E%7B2%7D%20%7D%7B2%7D%20%7D%20%3D%20%20%5Cdfrac%7Bv%7D%7B%20%5Csqrt%7B2%7D%20%7D%20%20%7D)
- Now, the vertical component of velocity of projectile at the height half of
will be –
![\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }](https://tex.z-dn.net/?f=%5Clongrightarrow%20%20%20%5Cbf%7Bv_%7B%28y%29%7D%3Dv_0%20%5C%3A%20sin%20%5Ctheta%20%7D)
![\longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }](https://tex.z-dn.net/?f=%20%5Clongrightarrow%20%5Cbf%7Bv_%7B%28y%29%7D%20%3D%20%5Cdfrac%7Bv%7D%7B%20%5Csqrt%7B2%7D%20%7D%20%20%5C%3A%20sin%20%5Ctheta%20%3D%20%20%5Cdfrac%7Bv%20%5C%3A%20sin%20%5C%3A%20%5Ctheta%7D%7B%20%5Csqrt%7B2%7D%20%7D%20%20%7D)
Therefore, the vertical component of velocity of projectile at this height will be–
☀️![\qquad](https://tex.z-dn.net/?f=%5Cqquad)
![\pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }](https://tex.z-dn.net/?f=%20%5Cpink%20%7B%5Cbf%7B%20%5Cdfrac%7Bv%20%5C%3A%20sin%20%5C%3A%20%20%5Ctheta%7D%7B%20%5Csqrt%7B2%7D%20%7D%7D%20%7D)
Answer:
a. 4.9 m
Explanation:
To solve this problem we must take into account that power is defined as the relationship between the work and the time in which the work is done.
P = W/t
where:
P = power = 95 [W] (units of watts)
W = work [J] (units of Joules)
t = time = 6.2 [s]
We can clear the work from the previous equation.
W = P*t
W = 95*6.2 = 589 [J]
Now we know that the work is defined by the product of the force by the distance, therefore we can express the work done with the following equation.
W = F*d
where:
F = force = 120 [N] (units of Newtons)
d = distance [m]
d = W/F
d = 589/120
d = 4.9 [m]
High frequency for the range of radio frequency electronmagnetic waves between 3 and 30 megahertz (MHz)