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3 years ago

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:

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

You might be interested in

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

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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