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anastassius [24]

3 years ago

9

A father is trying to teach his child to ice skate. As the child stands still, the father pushes him forward with an acceleratio

n of 2.0 m/s2 north. If the child's mass is 20 kg what is the force with which the father is pushing

1 answer:

rjkz [21]3 years ago

4 0

Hi there!

We can use Newton's Second Law:

$\Sigma F = ma$

∑F = net force (N)

m = mass (kg)

a = acceleration (m/s²)

We are given the mass and acceleration, so:

∑F = 20 · 2 = <u>40 N</u>

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{WORTH BRAINLIEST WITH AN EXPLANATION} Janine is a police officer patrolling a street. She drives 2.2 km east, then 4.4 km west,

icang [17]

Answer:

Options (3) and (6)

Explanation:

Janine drives 2.2 km East.

Since, distances measured in the East are positive,

Displacement = 2.2 km

Then she drives 4.4 km west

Displacement = -4.4 km

Followed by 1.7 km in the East

Displacement = 1.7 km

Total displacement = 2.2 - 4.4 + 1.7 = -0.5 km

$\overrightarrow{d}=-0.5$ km

Total distance covered by Janine = 2.2 + 4.4 + 1.7 = 8.3 km

d = 8.3 km

Therefore, Options (3) and (6) will be the answer.

5 0

3 years ago

Two metal spheres of identical mass m = 4.20 g are suspended by light strings 0.500 m in length. The left-hand sphere carries a

Yanka [14]

Answer:

Explanation:

Answer:

0.632 m

Explanation:

let the equilibrium separation between the charges is d and the angle made by string with the vertical is θ.

According to the diagram,

d = L Sinθ + L Sinθ = 2 L Sinθ .....(1)

Let T be the tension in the string.

Resolve the components of T.

T Sinθ = k q1 q2 / d^2

T Sinθ = k q1 q2 / (2LSinθ)² .....(2)

T Cosθ = mg .....(3)

Dividing equation (2) by equation (3), we get

tanθ = k q1 q2 / (4 L² Sin²θ x mg)

tan θ Sin²θ = k q1 q2 / (4 L² m g)

For small value of θ, tan θ = Sin θ

So,

Sin³θ = k q1 q2 / (4 L² m g)

Sin³θ = (9 x 10^9 x 0.785 x 10^-6 x 1.47 x 10^-6) / (4 x 0.5 x 0.5 x 4.20 x 10^-3 x 9.8)

Sin³θ = 0.2523

Sinθ = 0.632

θ = 39.2 degree

So, the separation between the two charges, d = 2 x L x Sin θ

d = 2 x 0.5 x 0.632 = 0.632 m

6 0

3 years ago

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at In a few mi

Nataliya [291]

Answer:

2274 J/kg ∙ K

Explanation:

The complete statement of the question is :

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.

$m_{m}$ = mass of metal = 400 g

$c_{m}$ = specific heat of metal = ?

$T_{mi}$ = initial temperature of metal = 100 °C

$m_{a}$ = mass of aluminum cup = 100 g

$c_{a}$ = specific heat of aluminum cup = 900.0 J/kg ∙ K

$T_{ai}$ = initial temperature of aluminum cup = 15 °C

$m_{w}$ = mass of water = 500 g

$c_{w}$ = specific heat of water = 4186 J/kg ∙ K

$T_{wi}$ = initial temperature of water = 15 °C

$T$ = Final equilibrium temperature = 40 °C

Using conservation of energy

heat lost by metal = heat gained by aluminum cup + heat gained by water

$m_{m} c_{m} (T_{mi} - T) = m_{a} c_{a} (T - T_{ai}) + m_{w} c_{w} (T - T_{wi} ) \\(400) (100 - 40) c_{m} = (100) (900) (40- 15) + (500) (4186) (40 - 15)\\ c_{m} = 2274 Jkg^{-1}K^{-1}$

7 0

4 years ago

Which of these materials is permeable?

kogti [31]

Correct answer is letter B. sandstone

8 0

3 years ago

Two men, Joel and Jerry, each pushes an object that are identical on a horizontal frictionless floor starting from rest. Joel an

Nataliya [291]

Answer:

The work done by Joel is greater than the work done by Jerry.

Explanation:

Let suppose that forces are parallel or antiparallel to the direction of motion. Given that Joel and Jerry exert constant forces on the object, the definition of work can be simplified as:

$W = F\cdot \Delta s$

Where:

$W$ - Work, measured in joules.

$F$ - Force exerted on the object, measured in newtons.

$\Delta s$ - Travelled distance by the object, measured in meters.

During the first 10 minutes, the net work exerted on the object is zero. That is:

$W_{net} = W_{Joel} - W_{Jerry}$

$W_{net} = F\cdot \Delta s - F\cdot \Delta s$

$W_{net} = (F-F)\cdot \Delta s$

$W_{net} = 0\cdot \Delta s$

$W_{net} = 0\,J$

In exchange, the net work in the next 5 minutes is the work done by Joel on the object:

$W_{net} = W_{Joel}$

$W_{net} = F\cdot \Delta s$

Hence, the work done by Joel is greater than the work done by Jerry.

7 0

3 years ago