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3 years ago

15

A car was moving at 14 m/s After 30 s, its speed increased to 20 m/s. What was the acceleration during this time ( need help fas

t!!!)

1 answer:

Arada [10]3 years ago

7 0

Answer:let initial velocity u=14m/s

Final velocity v=20m/s

Time taken t=30

Acceleration =a

V=u +at

a= (20-14)/30

a=0.2m/s^2

Explanation:

Acceleration is the change in velocity with respect to time.

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What do Babys Come From

arlik [135]

Answer:

Mother

Explanation:

Is this actual question

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4 years ago

If the potential due to a point charge is 490 V at a distance of 10 m, what are the sign and magnitude of the charge?

uysha [10]

Answer:

+5.4×10⁻⁷ C

Explanation:

Electric potential: This can be defined as the work done in bringing a unit charge from infinity to that point against the action of the field. The S.I unit of potential is volt (V)

The formula for potential is

V = kq/r............................ Equation 1

Where V = electric potential, k = proportionality constant, q = charge, r = distance.

making q the subject of the equation,

q = Vr/k............................ Equation 2

Given: V = 490 V, r = 10 m,

Constant: k = 9×10⁹ Nm²/C²

Substitute into equation 2

q = 490(10)/(9×10⁹)

q = 5.4×10⁻⁷ C

q = +5.4×10⁻⁷ C

Hence the charge is +5.4×10⁻⁷ C

7 0

3 years ago

A small bolt with a mass of 33.0 g sits on top of a piston. The piston is undergoing simple harmonic motion in the vertical dire

densk [106]

Answer:

0.027m

Explanation:

the bolt loses contact with the piston only when acceleration due to gravity equals acceleration of piston

ω² * A = g where ω is angular velocity, A amplitude, g acceleration due to gravity

ω is given by 2πf, ω² is 4π²f²

A= g/4π²f² depending on the value of g used either 10m/s² or 9.8m/s²,

i used 10m/s² in this answer

5 0

4 years ago

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)

Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is $U = \int\limits^T_0 {P(t)} \, dt$

Compute U if the bulb remains on for 5h

Answer:

The value is $U = 7.563 *10^{5} \ J$

Explanation:

From the question we are told that

The power rating of the bulb is $P = 100 \ W$

The resistance is $R = 143 \ \Omega$

The voltage is $V = V_o sin [2 \pi ft]$

The energy expanded is $U = \int\limits^T_0 {P(t)} \, dt$

The voltage $V_o = 110 \ V$

The frequency is $f = 60 \ Hz$

The time considered is $t = 5 \ h = 18000 \ s$

Generally power is mathematically represented as

$P = \frac{V^2}{ R}$

=> $P = \frac{( 110 sin [2 \pi * 60t])^2}{ 144}$

=> $P = \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}$

So

$U = \int\limits^T_0 { \frac{ 110^2* [sin [120 \pi t])^2}{ 144}} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { ( sin^2 [120 \pi t]} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | T} \atop {0}} \right.$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | 18000} \atop {0}} \right.$

$U = \frac{110^2}{144} [\frac{18000}{2} - [\frac{1}{2} * \frac{sin(240 \pi (18000))}{240 \pi} ] ]$

=> $U = 7.563 *10^{5} \ J$

7 0

3 years ago

The leaves of a tree lose water to the atmosphere via the process of transpiration. A particular tree loses water at the rate of

Gnoma [55]

Answer:

The speed of the sap flowing in the vessel is 1.90 mm/s

Explanation:

Given:

The rate of water loss, Q = 3 × 10 ⁻⁸ m³/s

Number of vessels contained, n = 2000

Diameter of the vessel, D = 100 Mu m

thus, the radius of the vessel, r = 50 × 10⁻⁶ m

Now, the rate of flow is given as:

Q = AV .............(1)

where, A is the area of the cross-section

V is the velocity

Total area, A = n × (πr²)

substituting the values in the equation (1), we get

3 × 10 ⁻⁸ m³/s = [2000 × (π × (50 × 10⁻⁶)²)] × V

or

V = 1.909 × 10⁻³ m/s or 1.90 mm/s

Hence, the speed of the sap flowing in the vessel is 1.90 mm/s

7 0

3 years ago