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photoshop1234 [79]
2 years ago
6

You throw a tennis ball (mass 0.0570 kg) vertically upward. It leaves your hand moving at 15.0 m/s. Air resistance cannot be neg

lected, and the ball reaches a maximum height of 8.00 m. By how much does the total mechanical energy decrease from when the ball leaves your hand to when it reaches its maximum height? Express your answer with the appropriate units.
Physics
1 answer:
Deffense [45]2 years ago
3 0

Answer:195 J

Explanation:

Given

mass of ball m=0.0570\ kg

ball leaves the hand with u=15\ m/s

maximum height reached by ball h=8\ m

Initial Mechanical energy when ball just leaves the hand

M.E._1=(P.E.+K.E.)_1

M.E._1=(mgh)_1+(\frac{1}{2}mv^2)_1

considering hand to be datum so h_1=0[/tex]

so Potential energy at ground is zero

M.E._1=\frac{1}{2}\times m\times (15)^2

M.E._1=6.41\ J

Mechanical Energy at highest point

(M.E.)_2=(P.E.+K.E.)_2

at highest Point velocity is zero

(M.E.)_2=mgh_2+0

(M.E.)_2=0.0570\times 9.8\times 8

(M.E.)_2=4.46\ J

Decrease in Mechanical energy

(M.E.)_1-(M.E.)_2=6.41-4.46

(M.E.)_1-(M.E.)_2=1.95\ J

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The electric potential in a particular region of space varies only as a function of y-position and is given by the function V(y)
nikdorinn [45]

Answer:

E = 55.9583\ Volts/meter

Explanation:

First let's find the electric potential using y = 22.5:

V(y) = 1.69y^2 +15.6y+52.5

V(22.5) = 1.69(22.5)^2 + 15.6*22.5 + 52.5

V(22.5) = 1259.0625\ Volts

Then, to find the magnitude of the electric field, we just need to divide the electric potential by the distance y:

E = V/d

E = 1259.0625/22.5

E = 55.9583\ Volts/meter

3 0
2 years ago
A car moving with constant acceleration covers the distance between two points 60 m apart in 6.0 s. Its speed as it passes the s
BlackZzzverrR [31]

Answer:

The speed in the first point is: 4.98m/s

The acceleration is: 1.67m/s^2

The prior distance from the first point is: 7.42m

Explanation:

For part a and b:

We have a system with two equations and two variables.

We have these data:

X = distance = 60m

t = time = 6.0s

Sf = Final speed = 15m/s

And We need to find:

So = Inicial speed

a = aceleration

We are going to use these equation:

Sf^2=So^2+(2*a*x)

Sf=So+(a*t)

We are going to put our data:

(15m/s)^2=So^2+(2*a*60m)

15m/s=So+(a*6s)

With these equation, you can decide a method for solve. In this case, We are going to use an egualiazation method.

\sqrt{(15m/s)^2-(2*a*60m)}=So

15m/s-(a*6s)=So

\sqrt{(15m/s)^2-(2*a*60m)}=15m/s-(a*6s)

[\sqrt{(15m/s)^2-(2*a*60m)}]^{2}=[15m/s-(a*6s)]^{2}

(15m/s)^2-(2*a*60m)}=(15m/s)^{2}-2*(a*6s)*(15m/s)+(a*6s)^{2}

-120m*a=-180m*a+36s^{2}*a^{2}

0=120m*a-180m*a+36s^{2}*a^{2}

0=-60m*a+36s^{2}*a^{2}

0=(-60m+36s^{2}*a)*a

0=a1

\frac{60m}{36s^{2}} = a2

1.67m/s^{2}=a2

If we analyze the situation, we need to have an aceleretarion  greater than cero. We are going to choose a = 1.67m/s^2

After, we are going to determine the speed in the first point:

Sf=So+(a*t)

15m/s=So+1.67m/s^2*6s

15m/s-(1.67m/s^2*6s)=So

4.98m/s=So

For part c:

We are going to use:

Sf^2=So^2+(2*a*x)

(4.98m/s)^2=0^2+(2*(1.67m/s^2)*x)

\frac{24.80m^2/s^2}{3.34m/s^2}=x

7.42m=x

5 0
2 years ago
Why are we able to use nuclear energy for practical purposes? The reactions are controlled to regulate energy output. The reacti
musickatia [10]

Answer: The reactions are controlled to regulate energy output

Explanation:

Nuclear energy is produced when either two nuclei fuse to produce larger nucleus or a bigger nuclei splits into two smaller nucleus.

Uranium easily splits to produce nuclear energy. The reactions are controlled and energy output can be regulated. Nuclear energy in abundant yet controlled amount is available. Nuclear energy is mostly used in steam turbines to produce electricity.  

5 0
2 years ago
Read 2 more answers
Although blood cells are contained within a special liquid called plasma, the cells themselves are___________.
valentina_108 [34]

Answer:

Solid

Explanation:

The plasma is the liquid part of blood, it is 90% and accounts for 55% of blood volume. It is what red blood cells, white blood cells, and platelets move around in. These cells remain solid within the plasma. I hoped this helped!

8 0
2 years ago
A hungry 169169 kg lion running northward at 77.377.3 km/hr attacks and holds onto a 31.731.7 kg Thomson's gazelle running eastw
navik [9.2K]

Answer:  75,242.9 m/s

Explanation:

from the question we are given the following parameters

mass of Lion (ML) = 169,169 kg

velocity of lion (VL) = 777,377.7 m/s

mass of Gazelle (Mg) = 31,731.7 kg

velocity of Gazelle (Vg) = 63,863.8 kg

mass of Lion and Gazelle (M) = 200,900.7 kg

velocity of Lion and Gazelle (V) = ?

The first figure below shows the motion of the Lion and Gazelle with their direction.

The second diagram shows the motion of the Lion and Gazelle with their directions rearranged to form a right angle triangle.

from the triangle formed we can get the velocity of the Lion and Gazelle immediately after collision using their momentum and Phytaghoras theorem

momentum = mass x velocity

momentum of the Lion = 169,169 x 77,377.3 = 13,089,840,463.7 kgm/s

momentum of the Gazelle = 31,731.7 x 63,863.8 = 2,026,506,942.46 kgm/s

momentum of the Lion and Gazelle = 200,900.7  x V

now applying Phytaghoras theorem we have

13,089,840,463.7 + 2,026,506,942.46 =  200,900.7 x V

15,116,347,406.16 = 200,900.7 x V

V = 75,242.9 m/s

7 0
2 years ago
Read 2 more answers
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