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2 years ago

1. Mike wears an 8.50 kg backpack while scaling a cliff. After 90 minutes, Mike is 11.2 m above the starting point.

Physics

1 answer:

svet-max [94.6K]2 years ago

8 0

Answer:

a. 952Joules

b. 8008 joules

c. 1.48 watts

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A single-turn circular loop of wire of radius 45 mm lies in a plane perpendicular to a spatially uniform magnetic field. During

Lina20 [59]

Answer:

Magnitude of induced emf is 0.00635 V

Explanation:

Radius of circular loop r = 45 mm = 0.045 m

Area of circular loop $A=\pi r^2$

$A=3.14\times 0.045^2=0.00635m^2$

Magnetic field is increases from 250 mT to 350 mT

Therefore change in magnetic field $dB=250-350=100mT$

Emf induced is given by

$e=-N\frac{d\Phi }{dt}=-NA\frac{dB}{dt}$

$e=-0.00635\times \frac{100\times 10^{-3}}{0.10}=-0.00635V$

Magnitude of induced emf is equal to 0.00635 V

7 0

3 years ago

A square loop of side 7 cm is placed with the nearest side 2 cm from a long wire carrying a current that varies with time at a c

ioda

Answer:

Explanation:

side of the square loop, a = 7 cm

distance of the nearest side from long wire, r = 2 cm = 0.02 m

di/dt = 9 A/s

Integrate on both the sides

$\int _{0}^{i}di =9\int _{0}^{t}dt$

i = 9t

(a) The magnetic field due to the current carrying wire at a distance r is given by

$B = \frac{\mu_{0}i}{2\pi r}$

$B = \frac{\mu_{0}\times 9t}{2\pi r}$

(b)

Magnetic flux,

$\phi=\int B\times a dr$

$\phi=\int \frac{\mu_{0}\times 9t}{2\pi r}\times a dr$

$\phi=\frac{\mu_{0}\times 9t\times a}{2\pi}\times ln\left ( \frac{2 + 7}{2} \right )$

$\phi=\frac{\mu_{0}\times 9t\times 0.07}{2\pi}\times ln(4.5)$

$\phi = 1.89 \times 10^{-7}t$

(c)

R = 3 ohm

$e = -\frac{d\phi}{dt}$

magnitude of voltage is

e = 1.89 x 10^-7 V

induced current, i = e / R = (1.89 x 10^-7) / 3

i = 6.3 x 10^-8 A

8 0

3 years ago

Two insulated wires, each 2.64 m long, are taped together to form a two-wire unit that is 2.64 m long. One wire carries a curren

nikklg [1K]

Answer:

$4.77\ \text{A}$

Explanation:

F = Magnetic force = 4.11 N

$I_n$ = Net current

$I_2$ = Current in one of the wires = 7.68 A

B = Magnetic field = 0.59 T

$\theta$ = Angle between current and magnetic field = $65^{\circ}$

$l$ = Length of wires = 2.64 m

$I$ = Current in the other wire

Magnetic force is given by

$F=I_nlB\sin\theta\\\Rightarrow I_n=\dfrac{F}{lB\sin\theta}\\\Rightarrow I_n=\dfrac{4.11}{2.64\times 0.59 \sin65^{\circ}}\\\Rightarrow I_n=2.91\ \text{A}$

Net current is given by

$I_n=I_2-I\\\Rightarrow I=I_2-I_n\\\Rightarrow I=7.68-2.91\\\Rightarrow I=4.77\ \text{A}$

The current I is $4.77\ \text{A}$ .

8 0

2 years ago

A 38.5 kg man is in an elevator

Vika [28.1K]

I’m not too sure but I think it’s 8,91 m/s2

3 0

3 years ago

A box weighing 10 N is 4 m away from the fulcrum. How much force must be applied 1 m away on the opposite end of the fulcrum in

julia-pushkina [17]

Answer: Socratic

Explanation: i’m not sure but you can use Socratic it’s a good app that helps

4 0

3 years ago