A roller coaster is stationary at the top of a ramp.

An object is hanging by a string from the ceiling of an elevator. the elevator is moving upward with a constant speed. what is t

Anarel [89]

Since the elevator is moving with a constant speed and not accelerating, the tension in the string is simply the normal, routine, everyday boring weight of the object. Since the elevator is moving with a constant speed and not accelerating, the tension in the string is simply the normal, routine, everyday boring weight of the object.

6 0

3 years ago

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The current supplied by a battery as a function of time is ) -(0.64 A)e-/(6.0 hr). What is the total number of electrons transpo

Paha777 [63]

Answer:

The total number of electrons is $8.6\times10^{22}$

(3) is correct option.

Explanation:

Given that,

The current equation is

$I=0.64 e^{\dfrac{-t}{6.0 hr}}$

We know that,

The formula of charge

$q=\int_{0}^{\infty}{I dt}$

$q=\int_{0}^{\infty}{0.64 e^{\dfrac{-t}{6.0 hr}}dt$

$q=0.64\int_{0}^{\infty}{e^{\dfrac{-t}{6.0 hr}}dt$

$q=0.64\int_{0}^{\infty}{e^{\dfrac{-t}{21600}}dt$

$q=0.64(21600e^{\dfrac{-t}{21600}})_{0}^{\infty}$

$q=0.64(0-21600)$

$q=0.64\times21600$

$q=13824\ C$

We need to calculate the number of electron

Using formula of charge

$q=ne$

$n=\dfrac{q}{e}$

$n=\dfrac{13824}{1.6\times10^{-19}}$

$n=8.6\times10^{22}$

Hence, The total number of electrons is $8.6\times10^{22}$

7 0

3 years ago

You put a bottle of soft drink in a refrigerator and leave it until its temperature has dropped 12.0K .

Bingel [31]

Answer: a) - 437.8° F, b) - 261°c.

Explanation: a) the kelvin and Fahrenheit temperature scale are related by the formulae below.

5 (°F - 32) = 9 (k - 273)

Where °F = temperature in Fahrenheit and k = temperature in kelvin.

For question A, k = 12.0, by substituting to have the value for °F, we have

5(°F - 32) = 9 ( 12 - 273)

5(°F - 32) = 9(-261)

5(°F - 32) = - 2349

°F - 32 = - 2349/5

°F - 32 = - 469.8

°F = - 469.8 + 32

°F = - 437.8

Question B

The centigrade and kelvin scale are related by the formulae below

°c = k - 273

Where °c = temperature in centigrade and k = temperature in kelvin =12

°c = 12 - 273

°c = - 261

3 0

3 years ago

A spherical bowling ball with mass m = 3.6 kg and radius R = 0.101 m is thrown down the lane with an initial speed of v = 8.7 m/

luda_lava [24]

Answer:

1) The magnitude of the angular acceleration = 67.92 rad/ $s^{2}$

2) Magnitude of the linear acceleration = 2.744 m/ $s^{2}$

3) How long does it take the bowling ball to begin rolling without slipping = 0.906 s

4) How long does it take the bowling ball to begin rolling without slipping = 6.75 m

5) the final velocity is 6.21 m/s

Explanation:

the given information :

Bowling mass m = 3.6 kg

Radius = 0.101 m

Initial speed $v_{0}$ = 8.7 m/s

Coefficient of kinetic friction μ = 0.28

1) he magnitude of the angular acceleration of the bowling ball is

F = m a

$F_{g}$ = μ N , N = m g

$F_{g}$ = μ m g

1) The magnitude of the angular acceleration of the bowling ball as it slides down the lane:

momen inersia of Bowling ball I = (2/5) m $R^{2}$

torque τ = I α

τ = F R

I α = F R

(2/5) m $R^{2}$ α = μ m g R

α = (5 μ g / 2R) μ g R

= (5 x 0.28 x 9.8/ 2 x 0.101)

= 67.92 rad/ $s^{2}$

2) Magnitude of the linear acceleration of the bowling ball as it slides down the lane

F = - $F_{g}$ , $F_{k}$ is the force of kinetic friction

m a = - μ m g, remove m

the magnitude of linear accelaration is

a = μ g

= (0.28) (9.8)

= 2.744 m/ $s^{2}$

3) The bowling ball takes time to begin rolling without slipping:

The linear speed, $v_{t}$ = $v_{0}$ - a t

$v_{t}$ = $v_{0}$ - μ g t

the angular speed, ω = ω0 + α t

ω = ω0 + (5 μ g/2R ) t

$v_{t}$ = ω R

$v_{0}$ - μ g t = ω0 R + (5 μ g/2R ) t R

7 μ g t/2 = $v_{0}$ + ω0 R

hence,

t = (2 $v_{0}$ + ω0 R)/ 7 μ g

ω0 = 0 (no initial spin), therefore

t = 2 $v_{0}$ / 7 μ g

= 2 x 8.7 / 7 (0.28) (9.8)

= 0.906 s

4) How long it takes for the bowling ball to begin rolling without slipping, S

S = $v_{0}$ t - (1/2) a $t^{2}$

= (8.7) (0.906) - (1/2) (2.744) $0.906^{2}$

= 6.75 m

5) The final velocity

$v_{t}$ = $v_{0}$ - a t

$v_{t}$ = 8.7 - (2.744) (0.906)

$v_{t}$ = 6.21 m/s

4 0

3 years ago

Two objects of different masses have momentum of equal, non-zero magnitude. Which object has more kinetic energy?.

Semmy [17]

If they have the same momentum, then

m₁ v₁ = m₂ v₂

If m₁ > m₂, then we must have v₁ < v₂ to preserve the equality.

The object with the larger speed - the second object - thus has more kinetic energy.

8 0

2 years ago