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tia_tia [17]
2 years ago
13

A man holds a rectangular card in front of and parallel to a plane mirror. In order for him to see the entire image of the card,

what is the least mirror area needed ?
Physics
1 answer:
UkoKoshka [18]2 years ago
3 0

Answer:

mark as the brainly olss

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An electric toaster is rated 1200 watts at 120 volts. what is the total electrical energy used to operate the toaster for 30 sec
Anarel [89]
Energy is calculated as power*time, so give the wattage of 1200 W (equivalent to 1200 Joules/second) and time of 30 seconds, multiplying these gives 36000 J or 36 kJ of electrical energy.

If electrical charge or current is needed: Power = voltage * current, so given the power of 1200 watts and voltage of 120 V, current is 1200 W / 120 V = 10 Amperes. Charge is calculated by multiplying 10 A*30 s = 300 C.
3 0
2 years ago
A rectangular block weighs 240 N. the area of the block in contact with the floor is 20 cm2.calculate the pressure on the floor(
maks197457 [2]

Answer:

12 N/cm²

Explanation:

From the question given above, the following data were obtained:

Weight (W) of block = 240 N

Area (A) = 20 cm²

Pressure (P) =?

Next, we shall determine the force exerted by the block. This can be obtained as follow:

Weight (W) of block = 240 N

Force (F) =.?

Weight and force has the same unit of measurement. Thus, we force applied is equivalent to the weight of the block. Thus,

Force (F) = Weight (W) of block = 240 N

Force (F) = 240 N

Finally, we shall determine the pressure on the floor as follow:

Force (F) = 240 N

Area (A) = 20 cm²

Pressure (P) =?

P = F/A

P = 240 / 20

P = 12 N/cm²

Therefore, the pressure on the floor is 12 N/cm².

7 0
2 years ago
Physics in motion unit 6a the nature of waves
mylen [45]
What’s the question?
7 0
2 years ago
A rocket moves upward, starting from rest with an acceleration of +29.4 for 3.98 s. it runs out of fuel at the end of the 3.98 s
topjm [15]
U = 0, initial upward speed
a = 29.4 m/s², acceleration up to 3.98 s
a = -9.8 m/s², acceleration after 3.98s

Let h₁ =  the height at time t, for t ≤ 3.98 s
Let h₂ =  the height at time t > 3.98 s

Motion for  t ≤ 3.98 s:
h₁ = (1/2)*(29.4 m/s²)*(3.98 s)² = 232.854 m
Calculate the upward velocity at t = 3.98 s
v₁ = (29.4 m/s²)*(3.98 s) = 117.012 m/s

Motion for t  > 3.98 s
At maximum height, the upward velocity is zero.
Calculate the extra distance traveled before the velocity is zero.
(117.012 m/s)² + 2*(-9.8 m/s²)*(h₂ m) = 0
h₂ = 698.562 m

The total height is
h₁ + h₂ = 232.854 + 698.562 = 931.416 m

Answer: 931.4 m (nearest tenth)

6 0
2 years ago
Read 2 more answers
An 80 kg skateboarder moving at 3 m/s pushes off with her back foot to move faster. If her velocity increases to 5 m/s, what is
Arturiano [62]
1) The kinetic energy of an object is given by:
K= \frac{1}{2}mv^2
where m is the object's mass and v its speed.

By using this equation, we find the initial kinetic energy of the skateboarder:
K_i= \frac{1}{2}(80 kg)(3 m/s)^2=360 J
and the final kinetic energy as well:
K_f= \frac{1}{2}(80 kg)(5 m/s)^2=1000 J

So, her change in kinetic energy is
\Delta K=K_f-K_i=1000 J-360 J=640 J

2) The work-energy theorem states that the work done to increase the speed of an object is equal to the variation of kinetic energy of the object:
W=\Delta K
Therefore, the work done by the skateboarder is
W=\Delta K=640 J
7 0
2 years ago
Read 2 more answers
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