Which of the following is an example of an electromagnetic wave?

The four-wheel-drive all-terrain vehicle has a mass of 385 kg with center of mass G2. The driver has a mass of 75 kg with center

jenyasd209 [6]

The coefficient of friction is missing and it has a value of μ = 0.4

Answer:

a = 3.924 m/s²

Explanation:

I've attached the kinematic free body diagram.

Taking the sum of all upward and downward forces,

EFy = 0;

N1 + N2 - m_p•g - m_v•g = 0

N1 + N2 = m_p•g + m_v•g

Where;

N1 and N2 are the normal reactions at the wheels

m_p is the mass of the driver

m_v is the mass of the vehicle

g is the acceleration due to gravity.

Plugging in the relevant values in the question,we obtain;

N1 + N2 = (385 + 75) x 9.81

N1 + N2 = 4512.6N - - - (eq1)

Now, taking sum of all horizontal forces;

EFx = (m_p + m_v) x a

So,

μ(N1 + N2) = (mp + mv) x a

Thus,

0.4(N1 + N2) = (385 + 75)a

0.4(N1 + N2) = 460a

N1 + N2 = 1150a

From eq(1),N1 + N2 = 4512.6N

Thus,

1150a = 4512.6N

a = 4512.6/1150

a = 3.924 m/s²

Therefore, the acceleration, a = 3.924 m/s²

7 0

3 years ago

A 55-kg woman is wearing high heels.

Grace [21]

Answer:

Pressure, $P=1.90\times 10^7\ Pa$

Explanation:

It is given that,

Mass of the woman, m = 55 kg

Diameter of the circular cross section, d = 6 mm

Radius, r = 3 mm = 0.003 m

Let P is the pressure exerted on the floor. It is equal to the force acting on woman per unit area. It is given by :

$P=\dfrac{F}{A}$

$P=\dfrac{mg}{\pi r^2}$

$P=\dfrac{55\times 9.8}{\pi (0.003)^2}$

$P=1.90\times 10^7\ Pa$

So, the pressure exerted on the floor is $1.90\times 10^7\ Pa$ . Hence, this is the required solution.

7 0

3 years ago

When an object reflects all the light waves that strike it looks white or black?

egoroff_w [7]

Light waves can be from any color, depends on what it is bouncing no or reflecting off of.

3 0

3 years ago

A potential difference of 3.00 nV is set up across a 2.00 cm length of copper wire that has a radius of 2.00 mm. How much charge

Anvisha [2.4K]

The number of charge drifts are 3.35 X 10⁻⁷C

<u>Explanation:</u>

Given:

Potential difference, V = 3 nV = 3 X 10⁻⁹m

Length of wire, L = 2 cm = 0.02 m

Radius of the wire, r = 2 mm = 2 X 10⁻³m

Cross section, 3 ms

charge drifts, q = ?

We know,

the charge drifts through the copper wire is given by

q = iΔt

where Δt = 3 X 10⁻³s

and i = $\frac{V}{R}$

where R is the resistance

R = $\frac{pL}{r^{2} \pi }$

ρ is the resistivity of the copper wire = 1.69 X 10⁻⁸Ωm

So, i = $\frac{\pi(r)^{2}V }{pL}$

q = $\frac{\pi(r^{2} )Vt }{pL}$

Substituting the values,

q = 3.14 X (0.02)² X 3 X 10⁻⁹ X 3 X 10⁻³ / 1.69 X 10⁻⁸ X 0.02

q = 3.35 X 10⁻⁷C

Therefore, the number of charge drifts are 3.35 X 10⁻⁷C

3 0

3 years ago

A 10 m long steel beam is accidentally dropped by a construction crane from a height of 4.89 m. The horizontal component of the

Otrada [13]

Answer:

e = 1.21 mV

Explanation:

given,

length of rod = 10 m

height of drop = 4.89 m

Earth’s magnetic field = 12.4 µT

acceleration of gravity = 9.8 m/s²

velocity of the beam

$v = \sqrt{2gh}$

$v = \sqrt{2\times 9.8 \times 4.89}$

v = 9.79 m/s

emf of the beam

e = B l v

e = 12.4 x 10⁻⁶ x 9.79 x 10

e = 1.21 x 10⁻³ V

e = 1.21 mV

4 0

3 years ago