Can someone help me with physics quiz?

zhuklara [117]

Answer: 37.981 m/s

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the ball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$x=52 m$ is the point where the ball strikes ground horizontally

$V_{o}$ is the ball's initial speed

$\theta=0$ because we are told the ball is thrown horizontally

$t$ is the time since the ball is thrown until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=120m$ is the initial height of the ball

$y=0$ is the final height of the ball (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Knowing this, let's start by finding $t$ from (2):

$0=y_{o}+V_{o}sin(0\°) t+\frac{gt^{2}}{2}$ (3)

$0=y_{o}+\frac{gt^{2}}{2}$

$t=\sqrt{\frac{-2 y_{o}}{g}}$ (4)

$t=\sqrt{\frac{-2 (120 m)}{-9.8m/s^{2}}}$ (5)

$t=4.948 s$ (6)

Then, we have to substitute (6) in (1):

$x=V_{o}cos(0\°) t$ (7)

And find $V_{o}$ :

$V_{o}=\frac{x}{t}$ (8)

$V_{o}=\frac{52 m}{4.948 s}$ (9)

$V_{o}=10.509 m/s$ (10)

On the other hand, since we are dealing with constant acceleration (due gravity) we can use the following equation to find the value of the ball's final velocity $V$ :

$V=V_{o} + gt$ (11)

$V=10.509 m/s + (-9.8 m/s^{2})(4.948 s)$ (12)

$V=-37.981 m/s$ (13) This is the ball's final velocity, and the negative sign indicates its direction is downwards.

However, we were asked to find the ball's final speed, which is the module of the ball's final vleocity vector. This module is always positive, hence the speed of the ball just before it strikes the ground is 37.981 m/s (positive).

5 0

3 years ago

What helped provide support for the Plate Tectonic theory in the 1960s?

skad [1K]

D. Mineral and fossil matches from tests done on different continents.

3 0

3 years ago

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Light of frequency 5*10^14hz liberates electron with energy 2.31*10^-19 joule from a certain surface .what is wave length of ult

stealth61 [152]

Answer: Light of frequency 5 x 1014 HZ liberates electrons with energy 2.3 x 10-19from a certain metallic surface.

Explanation:

5 0

2 years ago

A solid ball is released from rest and slides down a hillside that slopes downward at 65.0" from the horizontal

PilotLPTM [1.2K]

Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
X: mgsin65 - F = mAx 
Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) 
Moment about center of mass: 
Fr = Iα 
Now Ax = rα 
and F = umgcos65 
mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) 
umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) 
ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) 
ugcos65 = 0.4(gsin65 - ugcos65) 
1.4ugcos65 = 0.4gsin65 
1.4ucos65 = 0.4sin65 
u = 0.4sin65/1.4cos65 
u = 0.613

3 0

3 years ago

Which metals have similar chemical properties? Check all that apply.

Alika [10]

Answer:

yes

Explanation:

3 0

2 years ago

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