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

Two students are playing paddle ball with a 5 kg spongy ball. If the ball is thrown at the batter with a speed of 5 m/s and boun

fenix001 [56]

Answer:

75 kgm/s

Explanation:

Impulse: This can be defined as the product of mass and change in velocity. The S.I unit is kgm/s.

From the question,

I = m(v-u)................... Equation 1

Where I = impulse, m = mass, v = final velocity, u = initial velocity.

Let the direction of the initial velocity be the positive direction.

Given: m = 5 kg, v = -10 m/s (bounce off), u = 5 m/s.

Substitute into equation 1

I = 5(-10-5)

I = 5(-15)

I = -75 kgm/s.

The negative sign tells that the impulse act in the same direction as the final velocity of the ball

Hence,

I = 75 kgm/s

3 0

2 years ago

A 3.45-kg centrifuge takes 100 s to spin up from rest to its final angular speed with constant angular acceleration. A point loc

Dafna11 [192]

Answer:

(a) 18.75 rad/s²

(b) 14920.78 rev

Explanation:

(a)

First we find the acceleration of the centrifuge using,

a = (v-u)/t......................... Equation 1

Where v = final velocity, u = initial velocity, t = time.

Given: v = 150 m/s, u = 0 m/s ( from rest), t = 100 s

Substitute into equation 1

a = (150-0)/100

a = 1.5 m/s²

Secondly we calculate for the angular acceleration using

α = a/r..................... Equation 2

Where α = angular acceleration, r = radius of the centrifuge

Given: a = 1.5 m/s², r = 8 cm = 0.08 m

substitute into equation 2

α = 1.5/0.08

α = 18.75 rad/s²

(b)

Using,

Ф = (ω'+ω).t/2........................... Equation 3

Where Ф = number of revolution of the centrifuge, ω' = initial angular velocity, ω = Final angular velocity.

But,

ω = v/r and ω' = u/r

therefore,

Ф = (u/r+v/r).t/2

where u = 0 m/s (at rest), = 150 m/s, r = 0.08 m, t = 100 s

Ф = [(0/0.08)+(150/0.08)].100/2

Ф = 93750 rad

If,

1 rad = 0.159155 rev,

Ф = (93750×0.159155) rev

Ф = 14920.78 rev

6 0

2 years ago

An inline skater skates on a circular track 120.0 m in diameter at a tangential speed of 9.20 m/s. If the skater’s mass is 68.5

jok3333 [9.3K]

Answer:

The centripetal force acting on the skater is <u>48.32 N.</u>

Explanation:

Given:

Radius of circular track is, $R=120.0\ m$

Tangential speed of the skater is, $v=9.20\ m/s$

Mass of the skater is, $m=68.5\ kg$

We are asked to find the centripetal force acting on the skater.

We know that, when an object is under circular motion, the force acting on the object is directly proportional to the mass and square of tangential speed and inversely proportional to the radius of the circular path. This force is called centripetal force.

Centripetal force acting on the skater is given as:

$F_c=\frac{mv^2}{R}$

Now, plug in the given values of the known quantities and solve for centripetal force, $F_c$ . This gives,

$F_c=\frac{68.5\times (9.20)^2}{120.0}\\\\F_c=\frac{68.5\times 84.64}{120}\\\\F_c=\frac{5797.84}{120}\\\\F_c=48.32\ N$

Therefore, the centripetal force acting on the skater is 48.32 N.

3 0

2 years ago

In a certain region of space, a uniform electric field has a magnitude of 4.30 x 104 n/c and points in the positive x direction.

denis23 [38]

The magnetic force exerted by a field E to a charge q is given by F=Eq. In this case, F=4.30*10^4*(6.80mu C). 1mu C=10^-6C, so F=4.30*6.80=10^-2=0.29N. The direction is in the x direction, the direction that the field is applied because the charge is positive.

5 0

2 years ago

An overnight rainstorm has caused a major roadblock. Three massive rocks of mass m1=584 kg, m2=838 kg, and m3=322 kg have blocke

Elena-2011 [213]

Answer:

Force must be applied to m₁ to move the group of rocks from the road at 0.250 m/s² = 436 N

Explanation:

Total force required = Mass x Acceleration,

F = ma

Here we need to consider the system as combine, total mass need to be considered.

Total mass, a = m₁+m₂+m₃ = 584 + 838 + 322 = 1744 kg

We need to accelerate the group of rocks from the road at 0.250 m/s²

That is acceleration, a = 0.250 m/s²

Force required, F = ma = 1744 x 0.25 = 436 N

Force must be applied to m₁ to move the group of rocks from the road at 0.250 m/s² = 436 N

8 0

2 years ago