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3 years ago

Three multiple choice science questions.

Physics

1 answer:

Ratling [72]3 years ago

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A ball is hit 1 meter above the ground with an initial speed of 40 m/s. If the ball is hit at an angle of 30 above the horizonta

Vika [28.1K]

Answer:

$t_t=4.131\ s$

Explanation:

Given:

height above the horizontal form where the ball is hit, $y=1\ m$

angle of projectile above the horizontal, $\theta=30^{\circ}$

initial speed of the projectile, $u=40\ m.s^{-1}$

Firstly we find the vertical component of the initial velocity:

$u_y=u.\sin\theta$

$u_y=40\times \sin30^{\circ}$

$u_y=20\ m.s^{-1}$

During the course of ascend in height of the ball when it reaches the maximum height then its vertical component of the velocity becomes zero.

So final vertical velocity during the course of ascend:

$v_y=0\ m.s^{-1}$

Using eq. of motion:

$v_y^2=u_y^2-2g.h$ (-ve sign means that the direction of velocity is opposite to the direction of acceleration)

$0^2=20^2-2\times 9.8\times h$

$h=20.4082\ m$ (from the height where it is thrown)

Now we find the time taken to ascend to this height:

$v_y=u_y-g.t$

$0=20-9.8t$

$t=2.041\ s$

Time taken to descent the total height:

we've total height, $h'=h+y$ $=20.4082+1$

$h'=u_y'.t'+\frac{1}{2} g.t'^2$

during the course of descend its initial vertical velocity is zero because it is at the top height, so $u_y'=0\ m.s^{-1}$

$21.4082=0+4.9t'^2$

$t'=2.09\ s$

Now the total time taken by the ball to hit the ground:

$t_t=t'+t$

$t_t=2.09+2.041$

$t_t=4.131\ s$

3 0

3 years ago

PLEASE HELP ME ASAP PLEASE PLEASE!!!!!

Effectus [21]

Answer:bottom left

Explanation:

8 0

3 years ago

Read 2 more answers

A spring with a pointer attached is hanging next to a scale marked in millimeters. Three different packages are hung from the sp

SOVA2 [1]

solution;

$the expression for force applied on the spring due to the load is\\f=k\Delta x\\here,\Delta x is the extension in the spring due to appling force\\given three case as following\\110N=k(40-x_{o})----------1\\240N=k(60-x_{o})----------2\\w=k(30-x_{o})-------------3\\$ $To calculate the accrual length of the spring,solve th equation 1 and 2\\\frac{110N}{240N}=\frac{k(40-x_{o})}{k(60-x_{o})}\\0.458=\frac{k(40-x_{o})}{k(60-x_{o})}\\0.458(60mm-x_{o})=(40mm-x_{o})\\x_{o}(1-0.458)=(40-60(0.458))mm\\x_{o}\frac{12.52}{0.542}\\=23.1mm\\to calculate the force on the spring in case,\\solve the equation 1 and 2\\\frac{110}{w}=\frac{k(40-x_{o})}{k(60-x_{o})}\\\frac{110}{w}=\frac{(40mm-23.1mm)}{30mm-23.1mm}\\w=\frac{110}{2.45}=44.9N$

8 0

3 years ago

One system consists of two like charges (q1 and q2) separated by a distance r. A second system consists of two charges that are

olganol [36]

fyt as per the question the magnitude of two like charges are given as q1 and q2.the separation distance is given as r unit.hence the potential energy is given as- $P.E=+\frac{1}{4\pi\epsilon} \frac{q1*q2}{r^2}$

here the potential energy is positive which means the force between two charges is repulsive.the potential energy is maximum which indirectly denotes that the system is unstable.due to this repulsion the smaller charge may accelerate.

coming to the same charges of opposite nature i.e unlike charges-here the magnitude of charges are same and separation distance is also are,so the potential energy will be given as- $P.E=-\frac{1}{4\pi\epsilon}\frac{q1*q2}{r^2}$