All categories

2 years ago

The two major problems with most motor vehicles are that they burn fossil fuels and _____________.

Physics

2 answers:

Citrus2011 [14]2 years ago

8 0

Answer:

A. Create radioactive waste i believe

Explanation:

PIT_PIT [208]2 years ago

5 0

Answer:

The two major problems with most motor vehicles are that they burn fossil fuels and. Release greenhouse gases ( D)

You might be interested in

A metal sphere with radius R1 has a charge Q1. Take the electric potential to be zero at an infinite distance from the sphere.

Airida [17]

Answer:

Part A : E = $\frac{1}{4\pi}$ ε₀ Q₁/R₁² Volt/meter

Part B : V = $\frac{1}{4\pi}$ ε₀ Q₁/R₁ Volt

Explanation:

Given that,

Charge distributed on the sphere is Q₁

The radius of sphere is R

₁

The electric potential at infinity is 0

The space around a charge in which its influence is felt is known in the electric field. The strength at any point inside the electric field is defined by the force experienced by a unit positive charge placed at that point.

If a unit positive charge is placed at the surface it experiences a force according to the Coulomb law is given by

F = $\frac{1}{4\pi}$ ε₀ Q₁/R₁²

Then the electric field at that point is

E = F/1

E = $\frac{1}{4\pi}$ ε₀ Q₁/R₁² Volt/meter

Part B

The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against electric forces.

Thus, the electric potential at the surface of the sphere of radius R₁ and charge distribution Q₁ is given by the relation

V = $\frac{1}{4\pi}$ ε₀ Q₁/R₁ Volt

4 0

2 years ago

A 1.50-m cylinder of radius 1.10 cm is made of a complicated mixture of materials. Its resistivity depends on the distance x fro

MArishka [77]

Answer:

Resistance = 3.35* $10^{-4}$ Ω

Explanation:

Since resistance R = ρ $\frac{L}{A}$

whereas $\rho(x) = a + bx^2$

resistivity is given for two ends. At the left end resistivity is $2.25* 10^{-8}$ whereas x at the left end will be 0 as distance is zero. Thus

$2.25*10^{-8} = a + b(0)^2\\ 2.25*10^{-8} = a + 0 \\2.25*10^{-8} = a$

At the right end x will be equal to the length of the rod, so $x = 1.50\\8.50*10^{-8} = (2.25*10^{-8}) + ( b* (1.50)^2 )\\8.50*10^{-8} - (2.25*10^{-8}) = b*2.25\\\frac{6.25*10^{-8}}{2.25} = b\\b = 2.77 *10^{-8}$

Thus resistance will be R = ρ $\frac{L}{A}$

where A = π $r^2$

so,

$R = \frac{8.50*10^{-8} * 1.50}{3.14*(1.10*10^{-2})^2} \\R=3.35 * 10 ^{-4}$

6 0

2 years ago

An electron is moving in the vicinity of a long, straight wire that lies along the z-axis. The wire has a constant current of 8.

viktelen [127]

Answer:

The force that the wire exerts on the electron is $-4.128\times10^{-20}i-6.88\times10^{-20}j+0k$

Explanation:

Given that,

Current = 8.60 A

Velocity of electron $v= (5.00\times10^{4})i-(3.00\times10^{4})j\ m/s$

Position of electron = (0,0.200,0)

We need to calculate the magnetic field

Using formula of magnetic field

$B=\dfrac{\mu I}{2\pi d}(-k)$

Put the value into the formula

$B=\dfrac{4\pi\times10^{-7}\times8.60}{2\pi\times0.200}$

$B=0.0000086\ T$

$B=-8.6\times10^{-6}k\ T$

We need to calculate the force that the wire exerts on the electron

Using formula of force

$F=q(\vec{v}\times\vec{B}$

$F=1.6\times10^{-6}((5.00\times10^{4})i-(3.00\times10^{4})j\times(-8.6\times10^{-6}) )$

$F=(1.6\times10^{-19}\times3.00\times10^{4}\times(-8.6\times10^{-6}))i+(1.6\times10^{-19}\times5.00\times10^{4}\times(-8.6\times10^{-6}))j+0k$

$F=-4.128\times10^{-20}i-6.88\times10^{-20}j+0k$

Hence, The force that the wire exerts on the electron is $-4.128\times10^{-20}i-6.88\times10^{-20}j+0k$

5 0

2 years ago

A 500 kg motorcycle accelerates at a rate of 2 m/s .how much force was applied to the motorcycle?

Aleksandr [31]

Answer:

by using formula F=ma which is m stand for mass a stand for acceleration. so 500kg × 2 ms^-2

8 0

2 years ago

Read 2 more answers

A circular saw blade with radius 0.175 m starts from rest and turns in a vertical plane with a constant angular acceleration of

adelina 88 [10]

Answer:

x = 11.23 m

Explanation:

For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.

Let's reduce to SI system units

θ = 155 rev (2pi rad / rev) = 310π rad

α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²

Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)

w² = w₀² + 2 α θ

w =√ 2 α θ

w = √(2 4pi 310pi)

w = 156.45 rad / s

The relationship between angular and linear velocity

v = w r

v = 156.45 0.175

v = 27.38 m / s

In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive

y = $v_{oy}$ t - ½ g t²

As it leaves the highest point its speed is horizontal

y = 0 - ½ g t²

t = √ (-2y / g)

t = √ (-2 (-0.820) /9.8)

t = 0.41 s

With this time we calculate the horizontal distance, because the constant horizontal speed

x = vox t

x = 27.38 0.41

x = 11.23 m

5 0

2 years ago