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3 years ago

A 104 horse-power suv speeds up from 21 m/s to 29 m/s in 3 seconds, what is the mass of the suv?

Physics

1 answer:

AleksAgata [21]3 years ago

4 0

Answer:

Correct answer: m = 1,160 kg

Explanation:

Given:

P = 104 hp = 77.48 kW = 77.48 · 10³ W

V₁ = 21 m/s Initial speed (velocity)

V₂ = 29 m/s Final speed (velocity)

t = 3 seconds The time interval during which the work was performed

m = ?

The formula for calculating power is:

P = A / Δt

Since work is a measure of energy change in this case kinetic:

P = ΔEk / Δt = (Ek₂ - Ek₁) / Δt = (m V₂²/ 2 - m V₁²/ 2) / Δt

P = m · (V₂² - V₁²) / 2 · Δt ⇒ m = (2 · P · Δt) / (V₂² - V₁²)

m = (2 · 77.48 · 10³ · 3) / (29² - 21²) = 464.88 · 10³ / (841 - 441)

m = 464.88 · 10³ / 400 = 1.16 · 10³ = 1,160 kg

m = 1,160 kg

God is with you!!!

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A battery with an emf of 12.0 V shows a terminal voltage of 11.7 V when operating in a circuit with two lightbulbs, each rated a

wariber [46]

<h2>Answer:</h2>

0.46Ω

<h2>Explanation:</h2>

The electromotive force (E) in the circuit is related to the terminal voltage(V), of the circuit and the internal resistance (r) of the battery as follows;

E = V + Ir --------------------(a)

Where;

I = current flowing through the circuit

But;

V = I x Rₓ ---------------------(b)

Where;

Rₓ = effective or total resistance in the circuit.

<em>First, let's calculate the effective resistance in the circuit:</em>

The effective resistance (Rₓ) in the circuit is the one due to the resistances in the two lightbulbs.

Let;

R₁ = resistance in the first bulb

R₂ = resistance in the second bulb

Since the two bulbs are both rated at 4.0W ( at 12.0V), their resistance values (R₁ and R₂) are the same and will be given by the power formula;

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

=> R = $\frac{V^{2} }{P}$ -------------------(ii)

Where;

P = Power of the bulb

V = voltage across the bulb

R = resistance of the bulb

To get R₁, equation (ii) can be written as;

R₁ = $\frac{V^{2} }{P}$ --------------------------------(iii)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iii) as follows;

R₁ = $\frac{12.0^{2} }{4}$

R₁ = $\frac{144}{4}$

R₁ = 36Ω

Following the same approach, to get R₂, equation (ii) can be written as;

R₂ = $\frac{V^{2} }{P}$ --------------------------------(iv)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iv) as follows;

R₂ = $\frac{12.0^{2} }{4}$

R₂ = $\frac{144}{4}$

R₂ = 36Ω

Now, since the bulbs are connected in parallel, the effective resistance (Rₓ) is given by;

$\frac{1}{R_{X} }$ = $\frac{1}{R_1}$ + $\frac{1}{R_2}$ -----------------(v)

Substitute the values of R₁ and R₂ into equation (v) as follows;

$\frac{1}{R_X}$ = $\frac{1}{36}$ + $\frac{1}{36}$

$\frac{1}{R_X}$ = $\frac{2}{36}$

Rₓ = $\frac{36}{2}$

Rₓ = 18Ω

The effective resistance (Rₓ) is therefore, 18Ω

<em>Now calculate the current I, flowing in the circuit:</em>

Substitute the values of V = 11.7V and Rₓ = 18Ω into equation (b) as follows;

11.7 = I x 18

I = $\frac{11.7}{18}$

I = 0.65A

<em>Now calculate the battery's internal resistance:</em>

Substitute the values of E = 12.0, V = 11.7V and I = 0.65A into equation (a) as follows;

12.0 = 11.7 + 0.65r

0.65r = 12.0 - 11.7

0.65r = 0.3

r = $\frac{0.3}{0.65}$

r = 0.46Ω

Therefore, the internal resistance of the battery is 0.46Ω

5 0

3 years ago

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A coil has N turns enclosing an area of A. In a physics laboratory experiment, the coil is rotated during the time interval Δt f

anzhelika [568]

Answer:

$\phi_i = BA$

Explanation:

magnetic flux is the count of magnetic field lines passing through a given loop or area

As we know that magnetic flux is given by the formula

$\phi = \vec B. \vec A$

here we also know that magnetic field B and plane of the coil is perpendicular in initial position

So the area vector is always perpendicular to the plane of the coil

so the angle between magnetic field and area vector is parallel to each other and this angle would be zero

so magnetic flux of the coil initially we have

$\phi = BAcos0 = BA$

6 0

3 years ago

An upright object 2.80 cm tall is placed 16.0 cm away from the vertex of a concave mirror with a center of curvature of 24.0 cm.

horrorfan [7]

Answer:

f = 12 cm

Explanation:

<u>Center of Curvature</u>:

The center of that hollow sphere, whose part is the spherical mirror, is known as the ‘Center of Curvature’ of mirror.

<u>The Radius of Curvature</u>:

The radius of that hollow sphere, whose part is the spherical mirror, is known as the ‘Radius of Curvature’ of mirror. It is the distance from pole to the center of curvature.

<u>Focal Length</u>:

The distance between principal focus and pole is called ‘Focal Length’. It is denoted by ‘F’.

The focal length of the spherical (concave) mirror is approximately equal to half of the radius of curvature:

$f = \frac{R}{2}$

where,

f = focal length = ?

R = Radius of curvature = 24 cm

Therefore,

$f = \frac{24\ cm}{2}$

8 0

3 years ago

Sometimes balance point may not be obtained on the potentiometer wire why

scoundrel [369]

Solution
Let a cell of emf E be connected across the entire length L of a potentiometer wire . Now , if the balance point is obtained at a length l during measurement of an unknown voltage

.
The balance point is not on the potentiometer wire - this statement means that

. In that case ,
l > L
V > E

8 0

3 years ago

A woman pushes a 27 kg lawnmower at a steady speed. She exerts a 120 N force in a direction 35◦ below the horizontal. The accele

nikklg [1K]

The vertical force exerted on the lawn is 68.8 N downward

Explanation:

The vertical force exerted by the lawnmower on the lawn is equal to the vertical component of the force applied, therefore:

$F_y = F sin \theta$