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ss7ja [257]
3 years ago
12

A bodybuilder loads a bar with 550 Newton’s (125 pounds) of weight and pushes the bar over her head 10 times. Each time she lift

s the weight 0.5 meters. How much work did she do?
Physics
1 answer:
jeyben [28]3 years ago
7 0
W=(550)(0.5)(10)=2750 J
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In the figure calculates the acceleration of the block friction not today
prohojiy [21]

Answer:

A fan pushes hot air out of a vent and into a room. The hot air displaces cold air in the room, causing the cold air to move closer to the floor.

The hot air displacing the cold air is an example of  transfer by

Explanation:

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3 0
3 years ago
Two solenoids are nested coaxially such that their magnetic fields point in opposite directions. Treat the solenoids as ideal. T
Scorpion4ik [409]

Answer:

Part a)

v = 6.4 \times 10^3 m/s

Part b)

v = 2.44 \times 10^4 m/s

Explanation:

Magnetic field due to outer solenoid inside it is given as

B_1 = \mu_0 \frac{N}{L} i

B_1 = (4\pi \times 10^{-7})(\frac{545}{0.205})(4.53)

B_1 = 0.015 T

Now similarly we have

magnetic field due to inner solenoid on it axis is given as

B_2 = \mu_0 \frac{N}{L} i

B_2 = (4\pi \times 10^{-7})(\frac{331}{0.189})(1.49)

B_2 = 3.28\times 10^{-3} T

Part a)

Now if a proton is revolving around the common axis of two solenoids

then we will have

qv(B_1 - B_2) = \frac{mv^2}{R}

(1.6 \times 10^{-19})(15 - 3.28)\times 10^{-3} = \frac{1.66\times 10^{-27} v}{5.69 \times 10^{-3}}

v = 6.4 \times 10^3 m/s

Part b)

Now proton is revolving in the field of outer cylinder only

so again we have

qvB_1 = \frac{mv^2}{R}

(1.6 \times 10^{-19})(15)\times 10^{-3} = \frac{1.66\times 10^{-27} v}{16.9 \times 10^{-3}}

v = 2.44 \times 10^4 m/s

7 0
3 years ago
A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm. Assume that the walls act lik
tamaranim1 [39]

Answer:

A) 1×10^6N/C

B) 2×10^-13N

C) 1×10^-2v

Explanation:

For parallel plates,the electric field E is given by:

E=s/eo

Where s= surface charge density

E = 10^-5/8.85×10^-12

E= 1.13×10^6 approximately 1×10^6NC^-2

B) Na has a charge of 1.6×10^-19

F= q×E= (1.13×10^6) × (1.6×10^-19)

F= 1.8×10^-13 approximately 2×10^-13N

C) Potential difference ,V= E×d

d=10nm=10^-8

V= 1.13×10^6 ×10^-8

V = 1.13×10^-2 approximately 1x10^-2v

8 0
3 years ago
If the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circular path were reduced,
kherson [118]
In this particular case, where car is movig through curvature, so it is moving in circular motion, force acting on car is centripetal force which holds car not to fly out. Centripetal force is always ponted in the middle of circle. Here frictional force has role of centripetal force. If frictional force is to weak, car would fly out of curvutare.
7 0
4 years ago
Read 2 more answers
Help! 5 questions for 25 points? Seems fair Please help
earnstyle [38]

Answer:

ezrfdjhsetnhtjhsksdjghbksfneunaifghejbcifkikjayhr

Explanation:

4 0
3 years ago
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