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Archy [21]
3 years ago
15

A 2-kg box is pushed by a force of 4 N for 2 seconds. It has an initial velocity vo = 2 m/s to the right. NOTE: Since this probl

em gives the TIME ∆t traveled, FIRST look at momenta values and calculate the IMPULSE using I = F ∆t = ∆p. After the momenta values are calculated, then calculate the kinetic energy and net work values.
Physics
1 answer:
IRISSAK [1]3 years ago
7 0

Answer:

Kf= 36 J

W(net) = 32 J

Explanation:

Given that

m = 2 kg

F= 4 N

t= 2 s

Initial velocity ,u= 2 m/s

We know that rate of change of linear momentum is called force.

F= dP/dt

F.t = ΔP

ΔP = Pf - Pi

ΔP = m v  - m u

v= Final velocity

By putting the values

4 x 2 = 2 ( v - 2)

8 =  2 ( v - 2)

4 = v - 2

v= 6 m/s

The final kinetic energy Kf

Kf= 1/2 m v²

Kf= 0.5 x 2 x 6²

Kf= 36 J

Initial kinetic energy Ki

Ki = 1/2 m u²

Ki= 0.5 x 2 x 2²

Ki = 4 J

We know that net work is equal to the change in kinetic energy

W(net) = Kf - Ki

W(net) = 36 - 4

W(net) = 32 J

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A capacitor with plates separated by a distance d is charged to a potential difference ΔVC. All wires and batteries are disconne
Tom [10]

Answer:

Explanation:

Initial separation of plate = d

final separation = 2d

The capacitance of the capacitor will reduce from C to C/2 because

capacitance = ε A / d

d is distance between plates.

As the batteries are disconnected , charge on the capacitor becomes fixed .

Initial charge on the capacitor

= Capacitance x potential difference

Q = C ΔV

Final charge will remain unchanged

Final charge = C ΔV

Final capacitance = C/2

Final potential difference = charge / capacitance

= C ΔV /  C/2

= 2 ΔV

Potential difference is doubled after the pates are further separated.

6 0
3 years ago
A student is told to use 20.0 g of sodium chloride to make an aqueous solution that has a concentration of 10.0 g/L (grams of so
Stells [14]

Answer:

she should add solute to the solvent

Explanation:

Given data :

Mass of the sodium chloride, = 20.0 g

Concentration of the solution = 10 g/L

Volume of 20.0 g of sodium chloride = 7.50 mL

Now, from the concentration, we can conclude that for 10 g of sodium chloride volume of the solution is 1 L

thus, for 20 g of sodium chloride  volume of the solution is 2 L or 2000 mL

also,

Volume of solution = Volume of solute(sodium chloride) + volume of solvent (water)

thus,

2000 mL = 7.5 mL + volume of solvent (water)

or

volume of water = (2000 - 7.5) mL

or

volume of water = 1992.5 mL

or

volume of water = 199.25 L ≈ 199 L

6 0
3 years ago
Read 2 more answers
Can there be situation when velocity is constant but speed is not
klasskru [66]

Answer:

no I don’t think there can be so my answer is No.

Okay then yes sorry that I must have gotten it wrong before.

Explanation:

4 0
2 years ago
Read 2 more answers
If a cross country runner covers a distance of 347 meters in 134 seconds what is her average speed?
a_sh-v [17]
347÷134=2.589552239 meters per second

2.589552239×60= 155.3731343 meters per hour

155.3731343 meters per hour= 0.096544389701642 miles per hour

hopefully this was right.
6 0
3 years ago
A baseball is batted from a height of 1.09 m with a speed of
kobusy [5.1K]

(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(c) The distance off course the ball would be carried is 0.38 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s

<h3>Resultant speed of the ball and crosswind</h3>

V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Learn more about projectiles here: brainly.com/question/11049671

5 0
1 year ago
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