Ask question

Ask question

All categories

3 years ago

14

A car left skid marks 78 m long on a road as it slid to a stop. If the car's acceleration was -3.9 m/s2, what was its velocity b

efore it began to skid, to the nearest m/s?

1 answer:

spayn [35]3 years ago

7 0

It is 25 I’m sure of it

You might be interested in

If a basketball travles a distance of 4 meters in 5 seconds whst is the average speed

tensa zangetsu [6.8K]

Well you would use The equation Average Speed= Distance/Time

S=4/5

S= 0.8

5 0

4 years ago

Read 2 more answers

What type of Earth scientist would be interested in understanding volcanic eruptions on

Vinvika [58]

Answer:

Marine geologists learn about the rocks and geologic processes of the ocean basins.

3 0

3 years ago

Read 2 more answers

A horizontal spring is lying on a frictionless surface. One end of the spring is attaches to a wall while the other end is conne

JulsSmile [24]

Answer:

v_f = 1.05 m/s

Explanation:

From conservation of energy;

E_f = E_i

Thus,

(1/2)m(v_f)² + (1/2)I(ω_f)² + m•g•h_f + (1/2)k•(x_f)² = (1/2)m(v_i)² + (1/2)I(ω_i)² + m•g•h_i + (1/2)k•(x_i)²

This reduces to;

(1/2)m(v_f)² + (1/2)Ik(x_f)² = (1/2)k•(x_i)²

Making v_f the subject, we have;

v_f = [√(k/m)] * [√((x_i)² - (x_f)²)]

We know that ω = √(k/m)

Thus,

v_f = ω[√((x_i)² - (x_f)²)]

Plugging in the relevant values to obtain;

v_f = 17.8[√((0.068)² - (0.034)²)]

v_f = 17.8[0.059] = 1.05 m/s

3 0

3 years ago

An initially stationary electron is accelerated by a uniform 640 N/C Electric Field. a) Calculate the kinetic energy of the elec

bulgar [2K]

Answer:

(a) 1.298 * 10^(-4) J

(b) 5.82 * 10^6 m/s

Explanation:

Parameters given:

Electric field, E = 640 N/C

Distance traveled by electron, r = 15 cm = 0.15 m

Mass of electron, m = 9.11 * 10^(-31) kg

Electric charge of electron, q = 1.602 * 10^(-19) C

(a) The kinetic energy of the electron in terms of Electric field is given as:

K = (q² * E² * r²) / 2m

Therefore, Kinetic energy, K, is:

K = [(1.602 * 10^(-19))² * 640² * 0.15²] / [2 * 9.11 * 10^(-31)]

K = {23651.981 * 10^(-38)} / [18.22 * 10^(-31)]

K = 1298.13 * 10^(-7) J = 1.298 * 10^(-4) J

(b) To find the final velocity of the electron, we have to first find the acceleration of the electron. This can be gotten by using the equations of force.

Force is generally given as:

F = ma

Electric force is given as:

F = qE

Therefore, equating both, we have:

ma = qE

a = (qE) / m

a = (1.602 * 10^(-19) * 640) / (9.11 * 10^(-31))

a = 112.54 * 10^(12) m/s² = 1.13 * 10^(14) m/s²

Using one of the equations of motion, we have that:

v² = u² + 2as

Since the electron started from rest, u = 0 m/s

Therefore:

v² = 2 * 1.13 * 10^(14) * 0.15

v² = 3.39 * 10^(13)

v = 5.82 * 10^6 m/s

The velocity of the electron after moving a distance of 15 cm is 5.82 * 10^6 m/s.

8 0

3 years ago

someone help pls. Two students, Mia and Peter, leave school to meet at the local coffee shop. Peter decides to jog to the coffee

cluponka [151]

Answer:

1) The distance further it takes Peter to arrive at the Coffee shop than Mia is 1.24 km

2) Mia's average speed is 6.00 km/hour

Peter's average speed is 8.48 km/hour

4) Mia's average velocity = Peter's average velocity = 6.00 km/hour

Explanation:

The given information from the diagram are;

The distance Peter jogs from school to the flower shop = 2.00 km

The distance Peter jogs from the Flower shop to the Coffee shop = 2.24 km.

The distance Mia walks from school directly to the Coffee shop = 3.00 km

The time it takes both Peter and Mia to arrive at the coffee shop = 30 minutes = 0.5 hour

1) The total distance Peter travels from school to the Coffee shop = 2.00 km + 2.24 km = 4.24 km

The distance Mia travels from school to the Coffee shop = 3.00 km

The distance further it takes Peter to arrive at the Coffee shop than Mia = 4.24 km - 3.00 km = 1.24 km

The distance further it takes Peter to arrive at the Coffee shop than Mia = 1.24 km

2) $Average \ speed = \dfrac{Total \ distance \ traveled}{Total \ time \ taken \ in \ the \ journey}$

$Therefore, \ Mia's \ average \ speed = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour$

Mia's average speed = 6.00 km/hour

$Peter's \ average \ speed = \dfrac{4.24 \ km}{0.5 \ hour}= 8.48 \ km/hour$

Peter's average speed = 8.48 km/hour

4) $Average \ velocicty = \dfrac{Displacement }{Time \ taken}$

The displacement from the School to the Coffee shop is 3.00 km for both Mia and Peter

The time it takes both Peter and Mia to arrive at the Coffee shop from the school is 30 minutes = 0.5 hour

$Therefore, \ Mia's \ average \ velocity = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour$

Mia's average velocity = 6.00 km/hour

$Peter's \ average \ velocity = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour$

Therefore, Peter's average velocity is also = 6.00 km/hour

6 0

3 years ago