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Xelga [282]
1 year ago
7

Which of the following devices would you expect to consume the most energy for each hour that it operates? a portable tape recor

der an electric space heater a desk lamp an electric can openerwhich of the following devices would you expect to consume the most energy for each hour that it operates? the answer is an electric space heater?
Physics
2 answers:
Alisiya [41]1 year ago
8 0

Answer: electric space heater

Explanation: The consumption of electrical energy per unit hour  is given by the power rating on the electrical appliance. The power rating depends on the electric current it withdraws to operate and the resistance of the appliance.

Portable tape recorder, a desk lamp and an electric can opener are small devices as compared to an electric space heater and with draw less current. thus, the power rating of the space heater is higher than other given appliances and it consumes more energy per hour.

Charra [1.4K]1 year ago
3 0
You are correct, the answer is an electric space heater.
The reason for this is because the heating element generates heat by resisting the flow of current. For this reason, other similar devices such as irons take up a large amount of energy as well.
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Eating 2500 Cal every day a friend of mine maintains a stable weight of 70 kg. One day, after eating 3500 Cal, he decided to do
Kaylis [27]

Answer:

Explanation:

Calories to be burnt = 3500 - 2500 = 1000 Cals .

Efficiency of conversion to mechanical work  is 25 % .

Work needed to burn this much of Cals = 1000 x 100 / 25 = 4000 Cals.

4000 Cals = 4.2 x 4000 = 16800 J  .

Work done in one jump = kinetic energy while jumping

= 1/2 m v²

= .5 x 70 x 3.3²

= 381.15 J .

Number of jumps required = 16800 / 381.15

= 44 .

4 0
1 year ago
Two satellites revolve around the Earth. Satellite A has mass m and has an orbit of radius r. Satellite B has mass 6m and an orb
melomori [17]

Answer:

aaaaa

Explanation:

M = Mass of the Earth

m = Mass of satellite

r = Radius of satellite

G = Gravitational constant

F=G\frac{Mm}{r^2}

F=G\frac{M6m}{r_b^2}

G\frac{Mm}{r^2}=G\frac{M6m}{r_b^2}\\\Rightarrow \frac{1}{r^2}=\frac{6}{r_b^2}\\\Rightarrow \frac{r_b^2}{r^2}=6\\\Rightarrow \frac{r_b}{r}=\sqrt{6}\\\Rightarrow r_b=2.44948r

r_b=2.44948r

8 0
2 years ago
A spaceship of frontal area 10 m2 moves through a large dust cloud with a speed of 1 x 106 m/s. The mass density of the dust is
Step2247 [10]

Answer:

The decelerating force is 3\times 10^{- 11}\ N

Solution:

As per the question:

Frontal Area, A = 10\ m^{2}

Speed of the spaceship, v = 1\times 10^{6}\ m/s

Mass density of dust, \rho_{d} = 3\times 10^{- 18}\ kg/m^{3}

Now, to calculate the average decelerating force exerted by the particle:

Mass,\ m = \rho_{d}V                                (1)

Volume, V = A\times v\times t

Thus substituting the value of volume, V in eqn (1):

m = \rho_{d}(Avt)

where

A = Area

v = velocity

t = time

m = \rho_{d}(A\times v\times t)                  (2)

Momentum,\ p = \rho_{d}(Avt)v = \rho_{d}Av^{2}t

From Newton's second law of motion:

F = \frac{dp}{dt}

Thus differentiating w.r.t time 't':

F_{avg} = \frac{d}{dt}(\rho_{d}Av^{2}t) = \rho_{d}Av^{2}

where

F_{avg} = average decelerating force of the particle

Now, substituting suitable values in the above eqn:

F_{avg} = 3\times 10^{- 18}\times 10\times 1\times 10^{6} = 3\times 10^{- 11}\ N

4 0
1 year ago
A 6000 kg lorry is reversing into a parking space at a speed of 0.5 m/s but collides with a car. The crumple zone of the car sto
zysi [14]

Answer:

3000 kg.m/s

Explanation:

Momentum, p is a product of mass and velocity hence

p=mv where m is mass and v is velocity.

Change in momentum is given by m(v_f-v_i) where subscripts f and i represent final and initial respectively. Since the lorry finally comes to rest then the final velocity is zero. Substituting the given figures then

Change in momentum= 6000(0-0.5)=-3000 kg.m/s

7 0
2 years ago
You are in a hot-air balloon that, relative to the ground, has a veloc- ity of 6.0 m/s in a direction due east. You see a hawk m
Ann [662]

Answer:

6.32 m/s 18.43° northeast

Explanation:

We express the velocity of hawk as:

v_{Hawk}=v_{balloon}+v_{HawkRelativetoBalloon}=6 x+2 y

We consider positive x towards east and positive y due north. So the magnitude is simply the square root of the square components:

|v_{hawk}|=\sqrt[]{6^2+2^2}=\sqrt{40}≈6.32 m/s

And the angle with respect to the east should be with:

arctan(\frac{2}{6} )=18.43 \°

8 0
2 years ago
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