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joja [24]
2 years ago
14

We can only tell the earth is rotating almost 960 miles an hour if we

Physics
1 answer:
alexira [117]2 years ago
8 0

Answer:

C

It is C i have to type this because i need 20 characters.

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The lowest point in Death Valley is 85 m below sea level. The summit of nearby Mt. Whitney has an elevation of 4420 m.What is th
mario62 [17]

Answer:

\Delta E=2.87\times 10^6\ J

Explanation:

It is given that,

Depth of Death valley is 85 m below sea level, h_i=-85\ m

The summit of nearby Mt. Whitney has an elevation of 4420 m, h_f=4420\ m

Mass of the hiker, m = 65 kg

We need to find the change in potential energy. It is given by :

\Delta E=mg(h_f-h_i)

\Delta E=65\times 9.8(4420-(-85))

\Delta E=2869685\ J

or

\Delta E=2.87\times 10^6\ J

So, the change in potential energy of the hiker is 2.87\times 10^6\ J. Hence, this is the required solution.

5 0
3 years ago
Please i need your help w/h/w you make the following measurements if an object:42kg, and 22m​
Rufina [12.5K]

Answer:

1.9 kg/m^3

Explanation:

The density of an object is given by

d=\frac{m}{V}

where

m is the mass of the object

V is its volume

In this problem,

m = 42 kg

V = 22 m^3

Substituting into the equation, we find the object's density:

d=\frac{42 kg}{22 m^3}=1.9 kg/m^3

8 0
2 years ago
1, 2, & 3.........................
oksian1 [2.3K]

Answer:

1 is correct

2 is -150  

4 0
3 years ago
Who may drive a vehicle in Texas?
nirvana33 [79]

Answer:

Yes

Explanation:

3 0
3 years ago
Read 2 more answers
A lumberjack (mass = 103 kg) is standing at rest on one end of a floating log (mass = 255 kg) that is also at rest. The lumberja
mina [271]

Answer:

(a) -1.18 m/s

(b) 0.84 m/s

Explanation:

(a)

The total linear momentum before the lumberjack begins to move is zero because all parts of the system are at res

From the law of conservation of momentum

m1v1+m2v2=0 hence m1v1=-m2v2 where m1 is mass of lumberjack, v1 is velocity of lumberjeck, m2 is mass of floating log, v2 is velocity of the floating log.

Substituting M1 for 103 Kg, V1 for 2.93 m/s, M2 for 255 Kg into the above equation we obtain

103Kg*2.93 m/s=-255Kg*V2

V2=-(103 kg*2.93 m/s)/255=-1.183490196  m/s

Hence V2=-1.18 m/s

(b)

For the second log

V(M1+M2)=m1v1 where V is the common velocity

V(103 Kg+255 Kg)=103 Kg*2.93 m/s

V=(103 Kg*2.93 m/s)/(103 Kg+255 Kg)=0.842988827  m/s

V=0.84 m/s

6 0
2 years ago
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