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2 years ago

Two students race up a flight of steps together. "Student A" reaches the top of the steps 2 seconds ahead of

Physics

1 answer:

oksano4ka [1.4K]2 years ago

3 0

Um student a because they were there a few seconds ahead

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An object is dropped from a height of 25 meters. At what velocity will it hit the ground? A. 7.0 meters/second B. 11 meters/seco

Basile [38]

$Final^{2}=Initial^{2}+2ad \\ x^{2}=0^{2}+2(9.8)(25) \\ x^{2}=490 \\ x=22.13 \\ C$

4 0

3 years ago

A boy throws a stone straight upward with an initial speed of 15m/s. What maximum height will the stone reach before falling bac

Vitek1552 [10]

$height=\frac{velocity^2}{2gravity}\\\\
velocity=15\frac{m}{s}\\\\
g=9,8\frac{m}{s^2}\\\\height=\frac{15^2}{2*9,8}=\frac{225}{19,6}\\\\
\boxed{height=11,48\ meters}$

6 0

2 years ago

An electron and a second particle both move in circles perpendicular to a uniformmagnetic field. The mass of the second particle

Katarina [22]

Answer:

The change on the second particle is $2.93\times 10^{-16}\ C$ .

Explanation:

The period of revolution of the particle in the magnetic field is given by the formula as follows :

$T=\dfrac{2\pi m}{Bq}$

It is given that the magnetic field is uniform. The mass of the second particle is the same as that of a proton but thecharge of this particle is different from that of a proton.

$m_s=m_p$

If both particles take the same amount of time to go once around their respective circles. So,

$T_e=T_s\\\\\dfrac{2\pi m_e}{Bq_e}=\dfrac{2\pi m_s}{Bq_s}\\\\\dfrac{m_e}{q_e}=\dfrac{m_p}{q_s}\\\\q_s=\dfrac{m_pq_e}{m_e}\\\\q_s=\dfrac{1.67\times 10^{-27}\times 1.6\times 10^{-19}}{9.11\times 10^{-31}}\\\\q_s=2.93\times 10^{-16}\ C$

So, the change on the second particle is $2.93\times 10^{-16}\ C$ .

7 0

2 years ago

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant (net) acceleration 29.4 m/s2 m /

Sidana [21]

Answer:

The maximum height is 2881.2 m.

Explanation:

Given that,

Acceleration = 29.4 m/s²

Time = 7.00 s

We need to calculate the distance

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Put the value into the formula

$s=0+\dfrac{1}{2}\times29.4\times7^2$

$s=720.3\ m$

We need to calculate the velocity

Using formula of velocity

$v=a\times t$

Put the value into the formula

$v=29.4\times7$

$v=205.8\ m/s$

We need to calculate the height

Using formula of height

$H=\dfrac{v^2}{2g}$

Put the value into the formula

$H=\dfrac{(205.8)^2}{2\times9.8}$

$H=2160.9\ m$

We need to calculate the maximum height

Using formula for maximum height

$H'=H+s$

Put the value into the formula

$H'=2160.9+720.3$

$H'=2881.2\ m$

Hence, The maximum height is 2881.2 m.

4 0

2 years ago

Explain the relationship between the current output of the power supply and the current through each component in the parallel c

baherus [9]

Explanation:

Current output at the battery will be current of entire circuit, while the current through each bulb in the parallel circuit is the total current circuit.

So, current output through power supply is i and current through each component be $i_1, i_2 , i_3$ considering only three component.

Then in a parallel circuit

$i = i_1+i_2+i_3$

4 0

2 years ago