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Furkat [3]
3 years ago
11

if a person gets eights hours of sleep every night but still feels exhausted in the morning he or she may suffer from

Physics
2 answers:
Monica [59]3 years ago
8 0
They suffer from being fatigued
sesenic [268]3 years ago
3 0

Answer:

depression

Explanation:

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The y-axis on a distance-time graph would be ____________. time distance speed
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Answer:

Distance

Explanation:

distance is in vertical axis,or y-axis and time is on the horizontal axis,or x-axis.

7 0
3 years ago
An inflated rubber balloon is rubbed with a wool cloth until an excess of 1.00 × 107 electrons is on the balloon. What is the ma
REY [17]

The magnitude of the charge on the balloon is 1.6 x 10⁻¹² C.

<h3>What is the magnitude of the charge on the ball?</h3>

The magnitude of the charge on the ball is calculated by determining the total charge equivalent to the given number of electrons.

The charge of one electron = 1.6 x 10⁻¹⁹ Coulombs

Now, we are going to estimated the total charge of 1 x 10⁷ electrons.

1 electron =  1.6 x 10⁻¹⁹ C

1 x 10⁷ electrons = ?

= (1 x 10⁷ electrons x 1.6 x 10⁻¹⁹ C) / (1 electron)

= 1.6 x 10⁻¹² C

Thus, the total charge of 1 x 10⁷ electrons is obtained by multiplying the magnitude of charge of one electron to the entire given electrons.

Learn more about charge of electron here: brainly.com/question/9317875

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8 0
1 year ago
 Will mark as Brainliest!! If possible, Please check these!!! Quiz and I need to pass!!! If you see anything wrong from my answe
jarptica [38.1K]
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3 years ago
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The square loop shown in the figure moves into a 0.80T magnetic field at a constant speed of 10m/s. The loop has a resistance of
PolarNik [594]
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3 years ago
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A pitcher throws a 0.140 kg baseball, and it approaches the bat at a speed of 35.0 m/s. The bat does Wnc = 75.0 J of work on the
Eva8 [605]

Answer:

The speed of the ball is 42.5 m/s

Explanation:

The initial kinetic energy of the ball is:

K_1=\frac{1}{2} m v_0^2=\frac{1}{2}*0.140 kg*(35.0 m/s)^2= 85.75 J

The speed of the ball after leaving the bat is:

K_2=K_1+W_{nc}\\ \frac{1}{2}mV^2= 85.75 J + 75 J\\ (\frac{1}{2}mV^2)2=( 160.75 J)2\\ mV^2= 321.5 J\\ V^2= \frac{321.5 J}{0.140kg} \\ V=\sqrt{\frac{321.5 J}{0.140kg}}

V=47.92 m/s

Using kinematic equation we can find the speed of the ball after being 25 m above the point of collision:

V_f^2-V^2=-2gh

V_f^2-(47.92 m/s)^2=-2*9.81m/s^2*25m

V_f^2=-2*9.81m/s^2*25m+(47.92 m/s)^2

V_f=\sqrt{-2*9.81m/s^2*25m+(47.92 m/s)^2}

V_f=42.5m/s

3 0
3 years ago
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