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3 years ago

Determine the net work a hiker must do on a 3.35-kg backpack to carry it up a

Physics

1 answer:

Sphinxa [80]3 years ago

8 0

Answer:

410.4J

Explanation:

Step one:

given

mass= 3.35kg

weight= 3.35*9.81= 32.86N

h=12.49m

Required

The net work done

Step two:

the work done is given as

WD= force* distance

WD= 32.86*12.49

WD= 410.4J

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ndicate whether the statement is true or false. Whenever we move, we alter the rate at which we move into the future. A. True B.

Ksju [112]

Whenever we move, we alter the rate at which we move into the future. This statement is true.

4 0

3 years ago

The density of aluminum is 2.70 g/ml. A piece of aluminum foil has a mass of 44 g. What is the volume of this piece of aluminum

anyanavicka [17]

Answer:

C) 16.3 ml

Explanation:

Density is equal to the ratio between the mass of an object and its volume:

$d=\frac{m}{V}$

where

m is the mass

V is the volume

In our problem, we know:

- density of aluminium: $d=2.70 g/mL$

- mass of the aluminium foil: $m=44 g$

So we can re-arrange the equation above and use these data to find the volume of the piece of aluminium foil:

$V=\frac{m}{d}=\frac{44 g}{2.70 g/mL}=16.3 mL$

7 0

3 years ago

Suppose the ring rotates once every 4.30 ss . If a rider's mass is 58.0 kgkg , with how much force does the ring push on her at

Stells [14]

Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

T = 4.30 s

Assuming the diameter of the ring equal to 16 m

radius, R = 8 m

$v = \dfrac{2\pi R}{T}$

$v = \dfrac{2\pi\times 8}{4.30}$

v = 11.69 m/s

now, Force does the ring push on her at the top

$- N - m g = \dfrac{-mv^2}{R}$

$N + m g = \dfrac{mv^2}{R}$

$N = \dfrac{mv^2}{R}- m g$

$N = m(\dfrac{v^2}{R}- g)$

$N = 58\times (\dfrac{11.69^2}{8}- 9.8)$

N = 422.36 N

The force exerted by the ring to push her is equal to 422.36 N.

6 0

3 years ago

What is the electric potential energy of an electron at the negative end of the cable, relative to the positive end of the cable

VashaNatasha [74]

Answer:

Electric potential energy at the negative terminal: $1.92\cdot 10^{-18}J$

Explanation:

When a particle with charge $q$ travels across a potential difference $\Delta V$ , then its change in electric potential energy is

$\Delta U = q \Delta V$

In this problem, we know that:

The particle is an electron, so its charge is

$q=-1.60\cdot 10^{-19}C$

We also know that the positive terminal is at potential

$V_+=0V$

While the negative terminal is at potential

$V_-=-12 V$

Therefore, the potential difference (final minus initial) is

$\Delta V = -12-0 = -12 V$

So, the change in potential energy of the electron is

$\Delta U = (-1.6\cdot 10^{-19})(-12)=1.92\cdot 10^{-18}J$

This means that the electron when it is at the negative terminal has $1.92\cdot 10^{-18}J$ of energy more than when it is at the positive terminal.

Since the potential at the positive terminal is 0, this means that the electric potential energy of the electron at the negative end is

$1.92\cdot 10^{-18}J$

3 0

3 years ago

What is a non contact force that attracts all objects to the centre of the earth

beks73 [17]

Answer:

Gravity. The weight-force or weight of an object is the force because of Gravity, which acts on the object attracting it towards the centre of the earth.

Hope this helps, 

 x

4 0

3 years ago