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2 years ago

A 110-pound person pulls herself up 4.0 feet. This took her 2.5 seconds. How much power was developed?

Physics

2 answers:

mrs_skeptik [129]2 years ago

5 0

Answer:

0.32 Hp is the answer. right

lesya [120]2 years ago

4 0

Answer: Power developed by her is <u><em>0.32 hp</em></u>

Explanation:

Work done , W= Force(F) X Displacement (d) in the direction of force

Thus work done , $W=Fd=110\times 4.0 ft-lbf = 440 ft-lbf$

Now 1.0 ft-lbf = 1.355 joules of energy

Therefore , work done, W= 440x 1.355 J = 596.2 J

Average Power , $P=\frac{work done(W)}{time (t)}$

=> $P=\frac{596.2}{2.5} watts = 238.5 watts$

We know 746 hp = 1.0 watts

=> $P=\frac{238.5}{746} hp=0.32 hp$

Thus average power developed by her is 0.32 hp

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A 2.5 m -long wire carries a current of 8.0 A and is immersed within a uniform magnetic field B⃗ . When this wire lies along the

leva [86]

Answer:

Explanation:

Let the magnetic field be B = B₁i + B₂j + B₃k

Force = I ( L x B ) , I is current , L is length and B is magnetic field .

In the first case

force = - 2.3 j N

L = 2.5 i

puting the values in the equation above

- 2.3 j = 8 [ 2.5 i x ( B₁i + B₂j + B₃k )]

= - 20 B₃ j + 20 B₂ k

comparing LHS and RHS ,

20B₃ = 2.3

B₃ = .115

B₂ = 0

In the second case

L = 2.5 j

Force = I ( L x B )

2.3i−5.6k = 8 ( 2.5 j x (B₁i + B₂j + B₃k )

= - 20 B₁ k + 20B₃ i

2.3i−5.6k = - 20 B₁ k + 20B₃ i

B₃ = .115

B₁ = .28

So magnetic field B = .28 i + .115 B₃

Part A

x component of B = .28 T

Part B

y component of B = 0

Part C

z component of B = .115 T .

8 0

2 years ago

A tennis ball bounces on the floor three times. If each time it loses 22.0% of its energy due to heating, how high does it rise

lesya692 [45]

Answer:

H = 109.14 cm

Explanation:

given,

Assume ,

Total energy be equal to 1 unit

Balance of energy after first collision = 0.78 x 1 unit

= 0.78 unit

Balance after second collision = 0.78 ^2 unit

= 0.6084 unit

Balance after third collision = 0.78 ^3 unit

= 0.475 unit

height achieved by the third collision will be equal to energy remained

H be the height achieved after 3 collision

0.475 ( m g h) = m g H

H = 0.475 x h

H = 0.475 x 2.3 m

H = 1.0914 m

H = 109.14 cm

6 0

2 years ago

The air in a pipe resonates at 150 Hz and 750 Hz, one of these resonances being the fundamental. If the pipe is open at both end

Xelga [282]

Answer:

Explanation:

Two frequencies with magnitude 150 Hz and 750 Hz are given

For Pipe open at both sides

fundamental frequency is 150 Hz as it is smaller

frequency of pipe is given by

$f=\frac{nv}{2L}$

where L=length of Pipe

v=velocity of sound

$f=150\ Hz$ for n=1

and f=750 is for n=5

thus there are three resonance frequencies for n=2,3 and 4

For Pipe closed at one end

frequency is given by

$f=\frac{(2n+1)}{4L}\cdot v$

for n=0

$f_1=\frac{v}{4L}$

$f_1=150\ Hz$

for n=2

$f_2=\frac{5v}{4L}$

Thus there is one additional resonance corresponding to n=1 , between $f_1$ and $f_2$

8 0

2 years ago

The potential (relative to infinity) at the midpoint of a square is 3.0 V when a point charge of +Q is located at one of the cor

Sveta_85 [38]

Answer:

d) 12 V

Explanation:

Due to the symmetry of the problem, the potential (relative to infinity) at the midpoint of the square, is the same for all charges, provided they be of the same magnitude and sign, and be located at one of the corners of the square.

We can apply the superposition principle (as the potential is linear with the charge) and calculating the total potential due to the 4 charges, just adding the potential due to any of them:

V = V(Q₁) + V(Q₂) +V(Q₃) + V(Q₄) = 4* 3.0 V = 12. 0 V

4 0

2 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

Lady bird [3.3K]

Answer:

The options are approximations of the exact answers:

A) $1\times10^6N/C$

B) $2\times10^{-13}N$

C) $1\times10^{-2}V$

D) Toward the inner wall

E) $3\times10^{-17}J$

Explanation:

A) The electric field in a parallel plate capacitor is given by the formula $E=\frac{\sigma}{k\epsilon_0}$ , where $\epsilon_0=8.85\times10^{-12}C/Vm$ and in our case $\sigma=10^5C/m^2$ and, for air, $k=1.00059$ , so we have:

$E=\frac{10^5C/m^2}{(1.00059)(8.85\times10^{-12}C/Vm)}=1.13\times10^6N/C$

B) The K+ ion has one elemental charge excess, so its charge is $q=1.6\times10^{-19}C$ , and the force a charge experiments under an electric field E is given by F=qE, so we have:

$F=(1.6\times10^{-19}C)(1.13\times10^6N/C)=1.81\times10^{-13}N$

C) The potential difference between two points separated a distance d under an uniform electric potential E is given by $\Delta V=dE$ , so we have:

$\Delta V=(10\times10^{-9}m)(1.13\times10^6N/C)=1.13\times10^{-2}V$

D) The electic field goes from positive to negative charges, so it goes towards the inner wall.

E) The work done by an electric field through a potential difference $\Delta V$ on a charge Q is $W=Q\Delta V$ , and is equal to the kinetic energy imparted on it, so we have:

$K=(3\times10^{-15}C)(1.13\times10^{-2}V)=3.39\times10^{-17}J$

5 0

2 years ago