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notsponge [240]

2 years ago

13

A ball on a string travels once around a circle with a circumference of 2.0 m. The tension in the string is 5.0 N. how much work

is done by tension?

1 answer:

SpyIntel [72]2 years ago

4 0

Answer:0

Explanation:

Given

circumference of circle is 2 m

Tension in the string $T=5 N$

$2\pi r=2$

$r=\frac{2}{2\pi }=\frac{1}{\pi }=0.318 m$

In this case Force applied i.e. Tension is Perpendicular to the Displacement therefore angle between Tension and displacement is $90^{\circ}$

$W=\int\vec{F}\cdot \vec{r}$

$W=\int Fdr\cos 90$

$W=0$

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6–23 an automobile engine consumes fuel at a rate of 22 l/h and delivers 55 kw of power to the wheels. if the fuel has a heating

Anna007 [38]

Explanation & answer:

Given:

Fuel consumption, C = 22 L/h

Specific gravity = 0.8

output power, P = 55 kW

heating value, H = 44,000 kJ/kg

Solution:

Calculate energy intake

E = C*P*H

= (22 L/h) / (3600 s/h) * (1000 mL/L) * (0.8 g/mL) * (44000 kJ/kg)

= (22/3600)*1000*0.8*44000 j/s

= 215111.1 j/s

Calculate output power

P = 55 kW

= 55000 j/s

Efficiency

= output / input

= P/E

=55000 / 215111.1

= 0.2557

= 25.6% to 1 decimal place.

8 0

2 years ago

If you were to triple the size of the Earth (R = 3R⊕) and double the mass of the Earth (M = 2M⊕), how much would it change the g

EastWind [94]

Answer:

Decreased by a factor of 4.5

Explanation:

"We have Newton formula for attraction force between 2 objects with mass and a distance between them:

$F_G = G\frac{M_1M_2}{R^2}$

where $G =6.67408 × 10^{-11} m^3/kgs^2$ is the gravitational constant on Earth. $M_1, M_2$ are the masses of the object and Earth itself. and R distance between, or the Earth radius.

So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:

$\frac{F_G}{f_g} = \frac{G\frac{M_1M_2}{R^2}}{G\frac{M_1m_2}{r^2}}$

$\frac{F_G}{f_g} = \frac{\frac{M_2}{R^2}}{\frac{m_2}{r^2}}$

$\frac{F_G}{f_g} = \frac{M_2}{R^2}\frac{r^2}{m_2}$

$\frac{F_G}{f_g} = \frac{M_2}{m_2}(\frac{r}{R})^2$

Since $M_2 = 2m_2$ and $r = R/3$

$\frac{F_G}{f_g} = \frac{2}{3^2} = 2/9 = 1/4.5$

So gravity would have been decreased by a factor of 4.5

8 0

2 years ago

The data table below shows the distribution of the energies of a pendulum 0.60 s into its motion. What is the missing value?

ohaa [14]

The total energy (also called mechanical energy) is the sum of the kinetic energy and potential energy:
$TE = KE + PE$
For this pendulum, we see that at t=0.60 s the total energy is TE=0.918 J while the potential energy is 0.054 J, so the kinetic energy (the missing value in the table) is
$KE=TE-PE=0.918 J - 0.054 J =0.864 J$

4 0

2 years ago

A silver wire 2.6 mm in diameter transfers a charge of 420 Cin 80 min. Silver contains 5.8 x 10^{28} free electrons per cubic me

never [62]

1) Current in the wire: 0.0875 A

The current in the wire is given by:

$I=\frac{Q}{t}$

where

Q is the charge passing a given point in the conductor

t is the time elapsed

In this problem, we have

Q = 420 C is the total charge passing through a given point in a time of

t = 80 min = 4800 s

So, the current is

$I=\frac{420 C}{4800 s}=0.0875 A$

2) Drift velocity of the electrons: $1.78\cdot 10^{-6} m/s$

The drift velocity of the electrons in the wire is given by:

$u = \frac{I}{nAq}$

where

I = 0.0875 A is the current

$n=5.8\cdot 10^{28}$ is the number of free electrons per cubic meter

A is the cross-sectional area

$q=1.6\cdot 10^{-19} C$ is the charge of one electron

The radius of the wire is

$r=\frac{d}{2}=\frac{2.6 mm}{2}=1.3 mm=0.0013 m$

So the cross-sectional area is

$A=\pi r^2=\pi (0.0013 m)^2=5.31\cdot 10^{-6} m^2$

So, the drift velocity is

$u = \frac{(0.0875 A)}{(5.8\cdot 10^{28})(5.31\cdot 10^{-6})(1.6\cdot 10^{-19}C)}=1.78\cdot 10^{-6} m/s$

4 0

2 years ago

A package is dropped from a helicopter that is descending steadily at a speed v0. After t seconds have elapsed, consider the fol

qaws [65]

Answer:

Part a)

$v = \sqrt{v_o^2 + g^2t^2}$

Part b)

$d = \frac{1}{2}gt^2$

Part c)

$v_f = v_o - gt$

Part d)

$d = \frac{1}{2}gt^2$

Explanation:

Part a)

As we know that speed of package is same as that of helicopter in horizontal direction

So after time "t" the velocity in x direction will remain constant while in Y direction it will go free fall

So we have

$v_y = -gt$

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{v_o^2 + g^2t^2}$

Part b)

Distance from helicopter is same as the distance of free fall

so we will have

$d = \frac{1}{2}gt^2$

Part c)

If helicopter is rising upwards with uniform speed

then final speed of the package after time t is given as

$v_f = v_i + at$

$v_f = v_o - gt$

Part d)

distance from helicopter

$d = \frac{1}{2}gt^2$

8 0

2 years ago

Read 2 more answers