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3 years ago

How much work would a child do while pulling a 12-kg wagon a distance of 4.3 m with a 22 n force?

Physics

2 answers:

Vesna [10]3 years ago

8 0

Answer:

Work:

"Work is distance,d traveled in the direction of the force,F which is applied to it."

Formula:

W=F.D, or we can place "s" instead of the "D".

So, now we have the data provided in the problem, as we have the;

mass,m= 12 kg,
distance,D=4.3 m,
force,F= 22 N.

Solution:

As putting the values inside the given formula;

W=(22).(4.3),
W=94.6 J(joules).⇒ The Answer.

lianna [129]3 years ago

5 0

The work done by the child while pulling the wagon is given by:

$W=Fd$

where F is the force applied and d is the distance through which the wagon has been moved.

In this problem, F=22 N and d=4.3 m, therefore the work done by the child is

$W=Fd=(22 N)(4.3 m)=94.6 J$

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. Egbert is in a rowboat four miles from the nearest point on Egbert a straight shoreline. He wishes to reach a house 12 miles f

swat32

Answer:

$t=\frac{13}{3} =4.33\ mi.hr^{-1}$

Explanation:

Given:

Distance from the shore, $s=4\ mi$

distance of house from the shore, $h=12\ mi$

speed of rowing, $v_r=3\ mi.hr^{-1}$

speed of walking, $v_w=4\ mi.hr^{-1}$

For the least amount of time to reach the house one must row at the nearest point on the shore and then walk from there.

Now the time taken to reach the shore:

$t_s=\frac{s}{v_s}$

$t_s=\frac{4}{3}\ mi.hr^{-1}$

Time taken in walking to house from the shore:

$t_h=\frac{h}{v_w}$

$t_h=\frac{12}{4}$

$t_h=3\ mi.hr^{-1}$

Therefore total time taken:

$t=t_h+t_s$

$t=\frac{13}{3} =4.33\ mi.hr^{-1}$

6 0

3 years ago

Suatu sistem gas berada didalam ruang yang fleksibel. Pada awalnya gas berada pada kondisi P1 = 1,5 × 10 pangkat 5 N/m2, T1 = 27

enot [183]

Answer:

16.00L

Explanation:

First you calculate the number of moles in the system:

$PV=nRT\\\\n=\frac{PV}{RT}\\\\n=\frac{(1.5*10^5N/m^2)(12L)}{(0.082L.atm/mol.K)(300.15K)}=73134.16\ mol$

To find the new volume of the system you use the following formula for an isobaric procedure:

$T_2-T_1=\frac{P}{nR}(V_2-V_1)\\\\V_2=\frac{nR}{P}(T_2-T_1)+V_1\\\\V_2=\frac{(73134.16\ mol)(0.082L.atm/mol.K)}{1.5*10^5N/m^2}(400.15-300.15)K+12L\\\\V_2=16.00L$

hence, the new volume is 16.00L

6 0

4 years ago

Assume you need to design a hydronic system that can deliver 80,000 Btu/hr. What flow rate of water is required if the temperatu

PolarNik [594]

Answer:

At 10°F change in temperature

Mass flowrate = 1.01 kg/s = 2.227 lbm/s

Volumetric flowrate = 1010 m³/s = 35667.8 ft³/s

At 20°F change in temperature

Mass flowrate = 0.505 kg/s = 1.113 lbm/s

Volumetric flowrate = 505 m³/s = 17833.9 ft³/s

Explanation:

80000 btu/hr = 23445.7 W

P = ṁc(ΔT)

ṁ = MASS flowrate

c = specific heat capacity of water = 4182 J/kg.K,

ΔT = change in temperature = 10°F

To convert, a change of 18°F is equal to a change of 10°C

A change of 10°F = 10×10/18 = 5.556°C = 5.556K

P = ṁc(ΔT)

23445.7 = ṁ(4182 × 5.556)

ṁ = 23445.7/(4182 × 5.556)

ṁ = 1.01 kg/s = 2.227 lbm/s

In volumetric flow rate, Q = density × mass flowrate = 1000 × 1.01 = 1010 m³/s = 35667.8 ft³/s

For a change of 20°F,

ΔT = change in temperature = 20°F

To convert, a change of 18°F is equal to a change of 10°C

A change of 20°F = 20×10/18 = 11.1111°C = 11.111K

P = ṁc(ΔT)

23445.7 = ṁ(4182 × 11.111)

ṁ = 23445.7/(4182 × 11.111)

ṁ = 0.505 kg/s = 1.113 lbm/s

In volumetric flow rate, Q = density × mass flowrate = 1000 × 0.505 = 505 m³/s = 17833.9 ft³/s

Hope this Helps!!!

4 0

4 years ago

the pull of gravity on mars is 3.7m/s^2. if a astronaut on mars lifts a 10 kg rock 1 m off the ground, just to see whats under i

Elanso [62]

Gravitational potential energy can be calculated using the formula:

$PE_{grav} =mgh$

Where:
PEgrav = Gravitational potential energy
m= mass
g = acceleration due to gravity
h = height

On Earth acceleration due to gravity is a constant 9.8 but since the scenario is on Mars, the pull of gravity is different. In this case, it is 3.7, so we will use that for g.

So put in what you know and solve for what you don't know.
m = 10kg
g = 3.7m/s^2
h = 1m

So we put that in and solve it.
$PE_{grav} =mgh$
$PE_{grav} =(10kg)(3.7m/s^{2})(1m)$
$PE_{grav} =37J$

7 0

4 years ago

Use the balanced equation of a nitrogen cycle pathway below to support the conservation of matter and energy.

mojhsa [17]

Explanation :

Given equation

$N_{2} + 8H \rightarrow 2NH_{3} + H_{2}$

This equation is balanced. because the number of reactants is equal to the number of products.

According to conservation of matter and energy,

The conservation of matter and energy states that matter can not create and can not destroy but conserved. we can only change one form to another form of the matter and energy.

So, we can say energy conserved and the ,mass of the system always same over time.

3 0

4 years ago