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REY [17]
3 years ago
10

How far can a mother push a 20.0 kg baby carriage, using a force of 62 N, if she can only do 2920 J of work? (Round to include t

he appropriate answer with only one number after the decimal like 2.1 or 2.0)
DO NOT INCLUDE UNITS BUT KNOW THAT DISTANCE IS IN METERS.
Physics
1 answer:
ICE Princess25 [194]3 years ago
4 0
2.3 would be the best answer Based on some calculator math i did
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The amount of water vapor in the air is known as the _______.
storchak [24]
It is C the humidity
5 0
3 years ago
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When hiking at altitude, the air pressure drops to 825,000Pa. Solve the force that site exerts against the side of a rectangular
OLga [1]

Answer:

The force is 1237500N

Explanation:

We can apply the following expression

pressure=\frac{Force}{Area}

To find the area of rectangular tent wall:

Area=length *width\\Area=2.00m*0.750m\\Area=1.5m^{2}

Now to find the force:

pressure=825,000pa

pressure=\frac{Force}{Area}

Force=pressure*Area\\Force=825000pa*1.5m^{2} \\Force=1237500N

7 0
3 years ago
For a given velocity of projection in a projectile motion, the maximum horizontal distance is possible only at ө = 45°. Substant
myrzilka [38]

First let us imagine the projectile launched at initial velocity V and at angle θ relative to the horizontal. (ignore wind resistance)

Vertical component y:

The initial vertical velocity is given as Vsinθ
The moment the projectile reaches the maximum height of h, the vertical velocity will be 0, therefore the time t taken to attain this maximum height is:

h = Vsinθ - gt
0 = Vsinθ - gt
t = (Vsinθ)/g

where g is  acceleration due to gravity

Horizontal component x:
The initial horizontal velocity is given as Vcosθ. However unlike the vertical component, this horizontal velocity remains constant because this is unaffected by gravity. The time to travel the horizontal distance D is twice the value of t times the horizontal velocity.
D = Vcosθ*[(2Vsinθ)/g] 
D = (2V²sinθ cosθ)/g 
 D = (V²sin2θ)/g

In order for D (horizontal distance) to be maximum, dD/dθ = 0
That is,

2V^2 cos2θ / g = 0
And since 2V^2/g must not be equal to zero, therefore cos(2θ) = 0
This is true when 2θ = π/2  or  θ = π/4


Therefore it is now<span> shown that the maximum horizontal travelled is attained when the launch angle is π/4 radians, or 45°.</span>

6 0
4 years ago
Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150
Phoenix [80]

Answer:

The temperature of the core raises by 2.8^{o}C every second.

Explanation:

Since the average specific heat of the reactor core is 0.3349 kJ/kgC

It means that we require 0.3349 kJ of heat to raise the temperature of 1 kg of core material by 1 degree Celsius

Thus reactor core whose mass is 1.60\times 10^{5}kg will require

0.3349\times 1.60\times 10^{5}kJ\\\\=0.53584\times 10^{5}kJ

energy to raise it's temperature by 1 degree Celsius in 1 second

Hence by the concept of proportionately we can infer 150 MW of power will increase the temperature by

\frac{150\times 10^{6}}{0.53584\times 10^{8}}=2.8^{o}C/s

5 0
3 years ago
A 2400W electric toaster is connected to a standard 120 V wall outlet. A family
jolli1 [7]

Answer:

1)  energy = 15.6 kWh,  2)    total_cost = $ 2.03

Explanation:

1) The energy dissipated is the product of the power and the time of use In a month it was used t = 6.5 h and the power of the toaster is

P = 2400 W = 2,400 kW

       energy = P  t

       energy = 2,400  6.5

        energy = 15.6  kWh

using rounding to a decimal

        energy = 15.6 kWh

2) The cost of energy is unit_cost = $ 0.13 / kWh

so the total cost

         total_cost = energy    unit_cost

         total_cost = 15.6   0.13

         total_cost = $ 2.028  

rounding to two decimal places

         total_cost = $ 2.03

5 0
3 years ago
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