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3 years ago

A ball is thrown upward at a velocity of 19.6 m/s. What is its velocity after 3.0, assuming negligible air resistance?

2 answers:

5 0

If the ball travels 19.6 meters every second and it travels for 3 seconds, obviously you have to multiply the velocity by 3:

19.6 x 3= 58.8m/s

MrRa [10]3 years ago

5 0

Answer:

After 3 seconds the velocity becomes 49 m/s

Explanation:

Applying upward displacement equation:

$Y = V_yt +\frac{1}{2}gt^2$

differentiate this equation,

$\frac{dy}{dt} = V_y + gt$

where; dy/dt is the velocity function;

After 3 seconds, the velocity becomes;

V = 19.6 + (9.8)(3)

V = 19.6 + 29.4

V = 49 m/s

Therefore, after 3 seconds the velocity becomes 49 m/s

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In the winter activity of tubing, riders slide down snow covered slopes while sitting on large inflated rubber tubes. To get to

goldenfox [79]

Answer:

17304 J

Explanation:

Complete statement of the question is :

In the winter activity of tubing, riders slide down snow covered slopes while sitting on large inflated rubber tubes. To get to the top of the slope, a rider and his tube, with a total mass of 84 kg , are pulled at a constant speed by a tow rope that maintains a constant tension of 350 N .

Part A

How much thermal energy is created in the slope and the tube during the ascent of a 30-m-high, 120-m-long slope?

Solution :

$T$ = tension force in the tow rope = 350 N

$L$ = length of the incline surface = 120 m

$W_{t}$ = work done by tension force = ?

The tension force acts parallel to incline surface, hence work done by tension force is given as

$W_{t} = T L\\W_{t} = (350) (120)\\W_{t} = 42000 J$

$h$ = height gained by the rider = 30 m

$m$ = total mass of rider and tube = 84 kg

Potential energy gained is given as

$U = mgh\\U = (84) (9.8) (30)\\U = 24696 J$

$Q$ = Thermal energy created

Using conservation of energy

$Q = W_{t} - U\\Q = 42000 - 24696\\Q = 17304 J$

7 0

3 years ago

5. Which unit of electricity does the work in a circuit?

Pachacha [2.7K]

Answer:

Volt

Explanation:

Voltage is what makes electric charges move. ... Voltage is also called, in certain circumstances, electromotive force (EMF). Voltage is an electrical potential difference, the difference in electric potential between two places. The unit for electrical potential difference, or voltage, is the volt.

The ohm is defined as an electrical resistance between two points of a conductor when a constant potential difference of one volt, applied to these points, produces in the conductor a current of one ampere, the conductor not being the seat of any electromotive force.

The coulomb (symbolized C) is the standard unit of electric charge in the International System of Units (SI). ... In terms of SI base units, the coulomb is the equivalent of one ampere-second. Conversely, an electric current of A represents 1 C of unit electric charge carriers flowing past a specific point in 1 s.

An ampere is a unit of measure of the rate of electron flow or current in an electrical conductor. One ampere of current represents one coulomb of electrical charge (6.24 x 1018 charge carriers) moving past a specific point in one second.

8 0

2 years ago

Can someone help me please??

Mrrafil [7]

1. Our solar system is the only place in the universe where gravity played a key part in the formation of planets.

2. Rocky planets are small, dense, and orbit relatively close to the sun, compared to the Jovian planets, which are large, less dense, and orbiting far from the sun.

3. _______

7 0

3 years ago

In an electric vehicle, each wheel is powered by its own motor. The vehicle weight is 4,000 lbs. By regenerative braking, its sp

Slav-nsk [51]

Answer:

the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

Explanation:

Given that;

weight of vehicle = 4000 lbs

we know that 1 kg = 2.20462

m = 4000 / 2.20462 = 1814.37 kg

Initial velocity $V_{i}$ = 60 mph = 26.8224 m/s

Final velocity $V_{f}$ = 30 mph = 13.4112 m/s

now we determine change in kinetic energy

Δk = $\frac{1}{2}$ m( $V_{i}$ ² - $V_{f}$ ² )

we substitute

Δk = $\frac{1}{2}$ ×1814.37( (26.8224)² - (13.4112)² )

Δk = $\frac{1}{2}$ × 1814.37 × 539.5808

Δk = 489500 Joules

we know that; 1 kilowatt hour = 3.6 × 10⁶ Joule

Δk = 489500 / 3.6 × 10⁶

Δk = 0.13597 ≈ 0.136 kWh

Therefore, the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

4 0

3 years ago

A skydiver of 75 kg mass has a terminal velocity of 60 m/s. At what speed is the resistive force on the skydiver half that when

ankoles [38]

Answer:

The speed of the resistive force is 42.426 m/s

Explanation:

Given;

mass of skydiver, m = 75 kg

terminal velocity, $V_T = 60 \ m/s$

The resistive force on the skydiver is known as drag force.

Drag force is directly proportional to square of terminal velocity.

$F_D = kV_T^2$

Where;

k is a constant

$k = \frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2}$

When the new drag force is half of the original drag force;

$F_D_2 = \frac{F_D_1}{2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_1}{2V_{T2}^2} \\\\\frac{1}{V_{T1}^2} = \frac{1}{2V_{T2}^2}\\\\2V_{T2}^2 = V_{T1}^2\\\\V_{T2}^2= \frac{V_{T1}^2}{2} \\\\V_{T2}= \sqrt{\frac{V_{T1}^2}{2} } \\\\V_{T2}= \frac{V_{T1}}{\sqrt{2} } \\\\V_{T2}= 0.7071(V_{T1})\\\\V_{T2}= 0.7071(60 \ m/s)\\\\V_{T2}= 42.426 \ m/s$

Therefore, the speed of the resistive force is 42.426 m/s

8 0

3 years ago