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Elenna [48]
3 years ago
10

A 0.54 kg bullfrog is sitting at rest on a level log. How large is the normal force of the log on the bullfrog?

Physics
1 answer:
RUDIKE [14]3 years ago
7 0

Answer:

<h2>5.3N</h2>

Explanation:

Step one:

given data

mass of bullfrog= 0.54kg

Step two:

Required

The force F in Newton

From newton first law

F=mg

The acceleration due to gravity acting on the frog is 9.81m/s^2

hence

F=0.54*9.81

F=5.3N

The normal force is 5.3N

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Convert the following to gram. a,250 kg b373 mg c,10 quanital,15 ton​
rusak2 [61]

Explanation:

250 KG = 250000 Grams

373 MG = 0.373 Grams

10 Quintals = 1000000 Grams

15 Ton = 15000000 Grams

Please Follow and inbox me

4 0
2 years ago
You place a 0.17 kg can of soup and a 0.31 kg jar of pickles on the kitchen counter, separated by a distance of 0.42 m. What is
tangare [24]

1.984 \times 10^{-11} \mathrm{N} \text { is the force of gravity exerted on the jar of pickles. }

<u>Explanation</u>:

According to Newton's third law that each force has an equal and opposite reaction force in this case both of the jars will exert the same force an each other

. The force is given by

\mathrm{F}=\frac{G \times M_{1} \times M_{2}}{r^{2}}

Where, F = force, G=\text { gravitational constant }=\left(6.67 \times 10^{-11}\right), mass \left(\mathrm{M}_{1}\right)=0.17 \mathrm{kg}, mass \left M_{2}= 0.31 \mathrm{kg} and Distance(r) = 0.42 m.

Substitute the values in the formula.

\mathrm{F}=\frac{6.67 \times 10^{-11} \times 0.17 \times 0.31}{0.42^{2}}

\mathrm{F}=\frac{3.51 \times 10^{-12}}{0.176}

\mathrm{F}=1.984 \times 10^{-11} \mathrm{N}

\text { The force of gravity exerted on the jar of pickles is } 1.984 \times 10^{-11} \mathrm{N} \text { . }

3 0
3 years ago
A car travels 45 km due north and 70 km west. What is the car's displacement? 6 points 24.7 km northeast 83.2 km northwest 76.5
HACTEHA [7]

Answer:

3150

Explanation:

if if you were two times 45 times 70 it would give you that answer

8 0
3 years ago
A 23.5 g piece of aluminum metal is initially at 100.0°C. It is dropped into a coffee cup-calorimeter containing 130.0 g of wate
vivado [14]

Answer: The molar heat capacity of aluminum is 25.3J/mol^0C

Explanation:

heat_{absorbed}=heat_{released}

As we know that,  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]         .................(1)

where,

q = heat absorbed or released

m_1 = mass of water = 130.0 g

m_2 = mass of aluminiunm = 23.5 g

T_{final} = final temperature = 26.0^oC=(273+26)K=299K

T_1 = temperature of water = 23^oC=(273+23)K=296K

T_2 = temperature of aluminium = 100^oC=273+100=373K

c_1 = specific heat of water= 4.184J/g^0C

c_2 = specific heat of aluminium= ?

Now put all the given values in equation (1), we get

130.0\times 4.184\times (299-296)=-[23.5\times c_2\times (299-373)]

c_2=0.938J/g^0C

Molar mass of Aluminium = 27 g/mol

Thus molar heat capacity =0.938J/g^0C\times 27g/mol=25.3J/mol^0C

5 0
3 years ago
An elevator (mass 4100 kg) is to be designed so that the maximum acceleration is 0.0400g. what is the maximum force the motor sh
melomori [17]

Answer:

The maximum force on the supporting cable is 80688 N.

The minimum force on the supporting cable is -164 N.

Explanation:

For maximum force movement of elevator is in upward direction. Thus, equation of motion is given by,

ma = T - mg

where m is the mass of elevator

a is acceleration of elevator

g is acceleration due to gravity

T is the maximum tension in the supporting cable

T = ma + mg

T = m (a + g)

T = 4100 ( 0.04g + 9.8)

T = 80688 N

This is the maximum force on the supporting cable.

For minimum force movement of elevator is in downward direction. Thus, equation of motion is given by,

ma = T - mg

where m is the mass of elevator

a= -0.04g is acceleration of elevator because elevator is moving downward

g is acceleration due to gravity

T is the minimum tension in the supporting cable

T = ma + mg

T = m (a + g)

T = 4100 ( 9.8 - 0.04g)

T = -164 N

This is the minimum force on the supporting cable.


7 0
3 years ago
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