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2 years ago

Please help me anyone i will mark brainliest

Physics

1 answer:

Mashcka [7]2 years ago

3 0

Answer:

for 4.567 I can't tell if it x $10^{3}$ or x $10^{5}$ so:

answer using x $10^{3}$ = 0.0000644697N

answer using x $10^{5}$ = 0.00644697

Explanation:

use equation F = GMm/ $R^{2}$

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The velocity of a diver just before hitting the water is -10.9 m/s, where the minus sign indicates that her motion is directly d

Volgvan

velocity of the diver is vertically downwards

given that

$v = -10.9 m/s$

time of motion of the diver

$t = 1.17 s$

so the displacement is given by

$d = v*t$

$d = -10.9 * 1.17$

$d = -12.75 m$

so it will be displaced downwards by 12.75 m

7 0

3 years ago

A ball thrown vertically upward is caught by the thrower after 2.00 s. Find (a) the initial velocity of the ball and (b) the max

ankoles [38]

Answer:

a) 9.8 m/s

b) 4.9 m

Explanation:

This problem is a good example of Vertical motion, where the main equations for this situation are:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

$V^{2}={V_{o}}^{2}-2gy$ (2)

Where:

$y$ is the height of the ball at a given time

$y_{o}=0m$ is the initial height of the ball (assuming the hand of the thrower the origin of the system)

$V_{o}$ is the initial velocity of the ball

$V$ is the final velocity of the ball

$t=2s$ is the time it takes for the ball to make the complete movement (from the moment it is thrown until it falls back into the pitcher's hands)

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

Knowing this, let's begin with the answers:

<h3>a) Initial velocity </h3>

In order to find the initial velocity $V_{o}$ of the ball, we will use equation (1) and $t=2s$ , taking into account that $y=0 m$ and $y_{o}=0m$ at this given time:

$0=0+V_{o}t-\frac{1}{2}gt^{2}$ (3)

Isolating $V_{o}$ :

$V_{o}=\frac{1}{2}gt$ (4)

$V_{o}=\frac{1}{2}(9.8 m/s^{2})(2 s)$ (5)

Then:

$V_{o}=9.8 m/s$ (6)

<h3>b) Maximum height </h3>

In this part, we will use equation (2), knowing the value of the height is maximum when $V=0$ . So, we will name this height as $y_{max}$ :

$0={V_{o}}^{2}-2gy_{max}$ (7)

Isolating $y_{max}$ :

$y_{max}=\frac{{V_{o}}^{2}}{2g}$ (8)

$y_{max}=\frac{{(9.8 m/s)}^{2}}{2(9.8 m/s^{2})}$ (9)

Finally:

$y_{max}=4.9 m$

4 0

3 years ago

A stone is dropped from the edge of a roof, and hits the ground with a velocity of -180 feet per second. How high (in feet) is t

Neko [114]

Answer:

d = 506.25 ft

Explanation:

As we know by kinematics that

$v_f^2 - v_i^2 = 2 a d$

here we know that initially the stone is dropped from rest from the edge of the roof

so here initial speed will be zero

now we have

$v_i = 0$

also the acceleration of the stone is due to gravity which is given as

$g = 32 ft/s^2$

now we have

$v_f = 180 ft/s$

so from above equation

$180^2 - 0 = 2(32)d$

$d = 506.25 ft$

6 0

3 years ago

What does a animal and a plant cell have in common.

Allushta [10]

Answer:

Structures that are common to plant and animal cells are the cell membrane, nucleus, mitochondria, and vacuoles. Structures that are specific to plants are the cell wall and chloroplasts.

Explanation:

if correct plzz brainliest :D

8 0

2 years ago

Read 2 more answers

An egg is thrown upward with a velocity of 4.5 m/s. How long will it take to reach it's maximum height?

Ray Of Light [21]

Answer:

0.45 seconds

Explanation:

Letting the value of g = 10 m/s/s

final velocity (v) = 0 m/s (since the egg will come to rest at the maximum height)

initial velocity(u) = 4.5 m/s

acceleration = -10 m/s/s (since the gravity is acting against the egg)

time = t seconds

From the first equation of motion:

t = 0.45 seconds

3 0

3 years ago