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3 years ago

How much force does it take to accelerate a 2,000-kg car at 4 m/s

Physics

1 answer:

Leona [35]3 years ago

3 0

Answer:

C. 8,000 N

Explanation:

Newton's second law of motion describes the relationship between an object's mass and the amount of force needed to accelerate it.

This law can be expresses as F=ma,

where

F - force acting on an object
m- the mass of an object
a - its acceleration

The more mass an object has, the more force you need to accelerate it. And the greater the force, the greater the object's acceleration.

F = ma

=2000×4 = 8,000 N

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What is the wavelength of light that is deviated in the first order through an angle of 13.1 ∘ by a transmission grating having

faltersainse [42]

Answer:

The wavelength of light is $4.53\times10^{-7}\ m$

Explanation:

Given that,

Angle = 13.1°

Number of slits = 5000

We need to calculate the wavelength of light

Diffraction of first order is defined as,

$d \sin\theta=n\lambda$ .....(I)

The separation of the slits

$d = \dfrac{1}{N}$

$d=\dfrac{1}{5000}$

$d=2\times10^{-6}\ m$

Now put the value in equation (I)

$2\times10^{-6}\sin13.1^{\circ}=\lambda$

Here, n = 1

$\lambda=4.53\times10^{-7}\ m$

Hence, The wavelength of light is $4.53\times10^{-7}\ m$

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3 years ago

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4 years ago

Read 2 more answers

A satellite, orbiting the earth at the equator at an altitude of 400 km, has an antenna that can be modeled as a 1.76-m-long rod

ivann1987 [24]

Answer:

The inducerd emf is 1.08 V

Solution:

As per the question:

Altitude of the satellite, H = 400 km

Length of the antenna, l = 1.76 m

Magnetic field, B = $8.0\times 10^{- 5}\ T$

Now,

When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:

$e = - l(vec{v}\times \vec{B})$

Here, velocity v is perpendicular to the rod

Thus

e = lvB (1)

For the orbital velocity of the satellite at an altitude, H:

$v = \sqrt{\frac{Gm_{E}}{R_{E}} + H}$

where

G = Gravitational constant

$m_{e} = 5.972\times 10^{24}\ kg$ = mass of earth

$R_{E} = 6371\ km$ = radius of earth

$v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}}{6371\times 1000 + 400\times 1000} = 7670.018\ m/s$

Using this value value in eqn (1):

$e = 1.76\times 7670.018\times 8.0\times 10^{- 5} = 1.08\ V$

5 0

3 years ago

Find the resistance of wire of 0.65m Radius 0.25 and resistivity 3x10-6 OHM

densk [106]

Complete Question:

Find the resistance of a wire of length 0.65 m, radius 0.25 mm and resistivity 3 * 10^{-6} ohm-metre.

Answer:

Resistance = 9.95 Ohms

Explanation:

Given the following data;

Length = 0.65 m

Radius = 0.25 mm to meters = 0.00025 m

Resistivity = 3 * 10^{-6} ohm-metre.

To find the resistance of the wire;

Mathematically, resistance is given by the formula;

$Resistance = P \frac {L}{A}$

Where;

P is the resistivity of the material.
L is the length of the material.
A is the cross-sectional area of the material.

First of all, we would find the cross-sectional area of the wire.

Area of circle = πr²

Substituting into the equation, we have;

Area = 3.142 * (0.00025)²

Area = 3.142 * 6.25 * 10^{-8}

Area = 1.96 * 10^{-7} m²

Now, to find the resistance of the wire;

$Resistance = 3 * 10^{-6} * \frac {0.65}{1.96 * 10^{-7}}$

$Resistance = 3 * 10^{-6} * 3316326.531$

Resistance = 9.95 Ohms

5 0

3 years ago

A total resistance of 3.03 Ω is to be produced by connecting an unknown resistance to a 12.18 Ω resistance. (a) What must be the

insens350 [35]

Answer:

(a) 4.0334Ω

(b)parallel

Explanation:

for resistors connected in parallel;

$\frac{1}{R_{eq} } =\frac{1}{R1}+\frac{1}{R2}$

Req =3.03Ω , R1 =12.18Ω

$\frac{1}{3.03 } =\frac{1}{12.18}+\frac{1}{R2}$

$\frac{1}{R2}=\frac{1}{3.03 }-\frac{1}{12.18}$

$\frac{1}{R2}=0.2479$

R2=1/0.2479

R2=4.0334Ω

(b)parallel connection is suitable for the desired total resistance. series connection can not be used to achieve a lower resistance as the equation for series connection is.

Req = R1+R2

3 0

3 years ago