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Dvinal [7]
3 years ago
9

What are the potential obstacles preventing you from completing your exercises as scheduled? How can you overcome those obstacle

s?
Physics
1 answer:
shusha [124]3 years ago
3 0

Answer:

Sleep, behavior patterns, mental state, and job

Explanation:

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A girl throws a tennis ball upward with an initial velocity of 4 m/s. What is the maximum displacement of the ball?
Jobisdone [24]

Answer:

0.82 m

Explanation:

The ball is in free fall - uniform accelerated motion with constant acceleration downward, a=g=-9.8 m/s^2 (acceleration of gravity). So we can use the following suvat equation to solve the problem:

v^2-u^2=2as

where

v is the final velocity

u = 4 m/s is the initial velocity

a is the acceleration

s is the displacement

At the maximum displacement, v = 0 (the velocity becomes zero). Substituting and solving for s, we find:

s=-\frac{u^2}{2a}=-\frac{4^2}{2(-9.8)}=0.82 m

8 0
3 years ago
. The magnitudes of two forces are measured to be 120 ± 5 N and 60 ± 3 N. Find the sum
Tju [1.3M]

Explanation:

120+60=180

120-60=60

5 0
2 years ago
Cars are safety tested to see how quickly they come to a stop under many conditions. The distance and time of one of these tests
stepladder [879]

Answer:Is C,0

Explanation:

5 0
3 years ago
Read 2 more answers
Kepler’s first law states that:
Likurg_2 [28]

Answer:

Kepler's first law means that planets move around the Sun in elliptical orbits. An ellipse is a shape that resembles a flattened circle. ... It is zero for a perfect circle.

6 0
2 years ago
Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103
son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

PE = -\frac{GMm}{R}

As M=m, then

PE = -\frac{Gm^2}{R}

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

U = -\frac{2Gm^2}{R}

dU = -\frac{2Gm^2}{R}

dKE = -dU

Therefore replacing we have that,

\frac{1}{2}mv^2 =\frac{Gm^2}{2R}

Re-arrange to find v,

v= \sqrt{\frac{Gm}{R}}

v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}

v = 816.7m/s

Therefore the  velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

2r = 2*10^3m

Therefore

dU = Gm^2(\frac{1}{R}-\frac{1}{2r})

v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}

v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}

v = 1.83*10^7m/s

Therefore the velocity when they are about to collide is 1.83*10^7m/s

7 0
3 years ago
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