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bekas [8.4K]
3 years ago
7

Tarzan, whose mass is 75 kg, is running from a cheetah. Tarzan is moving at 5 m/s when he grabs onto a hanging vine. How high of

f the ground does Tarzan swing ?
Physics
1 answer:
Fiesta28 [93]3 years ago
6 0

Answer:

Explanation:

His Kinetic energy = 1/2 m v^2

v = 5 m/s

m = 75 kg

Ke = 1/2 75 * 5^2

Ke = 937.5 Joules

This will be converted to PE when he reaches the maximum height he reaches. In other words KE = PE

PE = m * g * h

m = 75

g = 9.81

h = ?

PE = 937.5

937.5 = 75 * 9.81 * h

937.5 = 735.75 * h

937.5/735.75 = h

h= 1.27 meters

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How much heat is needed to change the temperature of 3 grams of gold (c = 0.129 ) from 21°C to 363°C? The answer is expressed to
Temka [501]

Q= mcΔT

Where Q is heat or energy

M is mass, c is heat capacitance and t is temperature

You have to convert Celsius into kelvin in order to use this formula I believe

Celsius + 273 = Kelvin

21 + 273 = 294K

363 + 273 = 636K

Now...

Q= (0.003)(0.129)(636-294)

Q= 0.132 J if you are using kilograms, in terms of grams which seems more appropriate the answer would be 132J of energy.  

3 0
3 years ago
Read 2 more answers
How much force is required to accelerate a 12 kg mass at 5 m/s 2
Savatey [412]

Answer:

60 N

Explanation:

This is just Newton's Second Law

F = m*a

F = ?

m = 12 kg

a = 5 m/^2

F = 5*12 = 60 Newtons

4 0
2 years ago
Water is contained in a closed, rigid 0.2 m 3 tank at an initial pressure of 5 bar and a quality of 50%. Heat transfer occurs un
elena55 [62]

Answer:

Final mass=0.89kg

Final pressure=5.6bar

Explanation:

To find mass,m=v/v1

But v1=vf + x(vg-vf)

Vf= 0.001093m^3/kg

Vg= 0.3748m^3/kg

V1= 0.001093+0.5(0.3748-0.001093)

V1= 0.225m^3/kg

M= 0.20/0.225 =0.89kg

Final pressure will be:

V/V1= P/P1

Cross multiply

VP1=V1P

P1= 0.225×5/0.2

P1=:5.6 bar

7 0
4 years ago
Someone please help !!
rodikova [14]
What is the Investigation about!
5 0
3 years ago
A very humble bumble bee is flying horizontally due North at a constant speed of 3.11 m/s. At the current location of the bumble
Reil [10]

To solve this problem we will apply the concepts of the Magnetic Force. This expression will be expressed in both the vector and the scalar ways. Through this second we can directly use the presented values and replace them to obtain the value of the magnitude. Mathematically this can be described as,

\vec{F_B} = q(\vec{V}\times \vec{B})

F_B = q|v||B| sin\theta

Here,

q = Charge

v = Velocity

B = Magnetic field

\theta = \text{Angle between } \vec{B} \text{ and } \vec{V}

Our values are given as,

\theta = 35.7\°

q = 22.5*10^{-9}C

B = 1.05*10^{-5}T

v = 3.11m/s

Replacing,

F_B = (22.5*10^{-9}C)(3.11 \times 1.05*10^{-5}) sin(35.1\°)

F_B = 4.224*10^{-13}N

Therefore the size of the magnetic force acting on the bumble bee is 4.22*10^{-13}N

3 0
3 years ago
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