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Pani-rosa [81]
2 years ago
6

The whale shark is the largest of all fish and can have the mass of three adult

Physics
1 answer:
ch4aika [34]2 years ago
3 0

Answer:

m = 20,000 kg

Explanation:

Force, F=2.5\times 10^4\ N

Acceleration of the shark, a=1.25\ m/s^2

It is required to find the mass of the shark. Let m is the mass. Using second law of motion to find it as follows :

F = ma

Putting the value of F and a to find m

m=\dfrac{F}{a}\\\\m=\dfrac{2.5\times 10^4}{1.25}\\\\m=20,000\ kg

So, the shark's mass is 20,000 kg.

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3 years ago
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water vapor contained in a piston–cylinder assembly undergoes an isothermal expansion at 240°c from a pressure of 7 bar to a pre
mafiozo [28]

The ideal gas constant is a proportionality constant that is added to the ideal gas law to account for pressure (P), volume (V), moles of gas (n), and temperature (T) (R). R, the global gas constant, is 8.314 J/K-1 mol-1.

According to the Ideal Gas Law, a gas's pressure, volume, and temperature may all be compared based on its density or mole value.

The Ideal Gas Law has two fundamental formulas.

PV = nRT, PM = dRT.

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V = Liters of Volume

n = Present Gas Mole Number

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M stands for Molar Mass of the Gas in grams Mol d for Gas Density in gL.

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7 0
10 months ago
Using newtons second law of motion, how fast for 100 KG object accelerates 350 N of force is applied to
fenix001 [56]

Answer:

3.5m/s^2

Explanation:

From Newton's second Law of Motion

F = ma

Where F is the applied force, m is the mass of the object and a is the acceleration.

F = 350 N

Mass = 100kg

350N = 100×a

a = 350/100

a = 3.5m/s^2

The acceleration of the object will be 3.5m/s^2

6 0
3 years ago
n a level football field a football is projected from ground level. It has speed 9.0 m/sm/s when it is at its maximum height. It
Pani-rosa [81]

Answer:

T = 0.225 s

Explanation:

The speed of a projectile at the highest point of its motion is the horizontal speed of the projectile. Considering the horizontal motion with negligible air resistance, we can use the following formula:

v_x = RT\\\\T = \frac{v_x}{R}

where,

T = Total time of ball in air = ?

R = Horizontal distance covered = 40 m

v_x = horizontal speed = 9 m/s

Therefore,

T = \frac{9\ m/s}{40\ m}

<u>T = 0.225 s</u>

4 0
2 years ago
Suppose someone pours 0.250 kg of 20.0ºC water (about a cup) into a 0.500-kg aluminum pan with a temperature of 150ºC. Assume th
Troyanec [42]

Answer : The temperature when the water and pan reach thermal equilibrium short time later is, 59.10^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of aluminium = 0.90J/g^oC

c_2 = specific heat of water = 4.184J/g^oC

m_1 = mass of aluminum = 0.500 kg = 500 g

m_2 = mass of water = 0.250 kg  = 250 g

T_f = final temperature of mixture = ?

T_1 = initial temperature of aluminum = 150^oC

T_2 = initial temperature of water = 20^oC

Now put all the given values in the above formula, we get:

500g\times 0.90J/g^oC\times (T_f-150)^oC=-250g\times 4.184J/g^oC\times (T_f-20)^oC

T_f=59.10^oC

Therefore, the temperature when the water and pan reach thermal equilibrium short time later is, 59.10^oC

8 0
3 years ago
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