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2 years ago

Jake has a car that accelerates at 10 m/s2. If the car has a mass of 1000 kg, how much force does the car produce?

Physics

2 answers:

aleksandr82 [10.1K]2 years ago

8 0

Answer:

10,000 Newtons

Please mark my answer.

Cheers, Jake

katen-ka-za [31]2 years ago

5 0

Answer:

10,000 N

Explanation:

Force = mass · acceleration

F = m · a

Jake's car:

Accelerates at 10 m/s²

Weighs 1000 kg

F = 1000 · 10

F = 10000 N

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A car driver traveling at a speed of 108km per hour ,sees a traffic light and stopped after travelling for 20seconds .Find the a

navik [9.2K]

Answer:

– 2.5 m/s²

Explanation:

We have,

• Initial velocity, u = 180 km/h = 50 m/s

• Final velocity, v = 0 m/s (it stops)

• Time taken, t = 20 seconds

We have to find acceleration, a.

$\longrightarrow$ a = (v ― u)/t

$\longrightarrow$ a = (0 – 50)/20 m/s²

$\longrightarrow$ a = –50/20 m/s²

$\longrightarrow$ a = – 5/2 m/s²

$\longrightarrow$ a = – 2.5 m/s² (Velocity is decreasing) [Answer]

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2 years ago

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nikdorinn [45]

Answer: D) Earthquake

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3 years ago

Read 2 more answers

A solenoid of radius 2.0 mm contains 100 turns of wire uniformly distributed over a length of 5.0 cm. It is located in air and c

tankabanditka [31]

Answer:

The magnetic field strength inside the solenoid is $5.026\times10^{-3}\ T$ .

Explanation:

Given that,

Radius = 2.0 mm

Length = 5.0 cm

Current = 2.0 A

Number of turns = 100

(a). We need to calculate the magnetic field strength inside the solenoid

Using formula of the magnetic field strength

Using Ampere's Law

$B=\dfrac{\mu_{0}NI}{l}$

Where, N = Number of turns

I = current

l = length

Put the value into the formula

$B=\dfrac{4\pi\times10^{-7}\times100\times2.0}{5.0\times10^{-2}}$

$B=0.005026=5.026\times10^{-3}\ T$

(b). We draw the diagram

Hence, The magnetic field strength inside the solenoid is $5.026\times10^{-3}\ T$ .

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3 years ago

The temperature of the cooling water as it leaves the hot engine of an automobile is 240 °F. After it passes through the radiato

djverab [1.8K]

Answer:571.09 kJ

Explanation:

Given

Temperature of cooling water from engine exit $=240^{\circ} F\approx 115.55^{\circ}C$

After Passing through the radiator its temperature decreases to $175^{\circ}F\approx 79.44^{\circ}F$

specific heat of water $=4.184 J/g^{\circ}C$

Volume of water $= 1 gallon\approx 3.78 L$

density of water $\rho =1 gm/mL$

Thus mass of water $=\rho \times V=3.78\times 1=3.78 kg$

Heat transferred to the surrounding is equal to heat absorbed by cooling water

$Q=m\cdot c\cdot \Delta T$

$Q=3.78\times 4.184\times 1000\times (115.55-79.44)$

$Q=3.78\times 4.184\times 1000\times (36.11)$

$Q=571.09 kJ$

4 0

3 years ago

Calculate the workdone to stretch an elastic string by 40cm if a force of 10 newton produces an extension of 4cm in it

notka56 [123]

Answer:

Workdone = 20 Joules

Explanation:

Given the following data;

Force = 10N

Extension, e = 4cm to meters = 4/100 = 0.04 meters

Workdone extension = 40cm to meters = 40/100 = 0.4 meters

To find the work done;

First of all, we would find the spring constant using the formula;

Force = spring constant * extension

10 = spring constant * 0.04

Spring constant = 10/0.04

Spring constant = 250 N/m

Next, we find the work done;

Workdone = ½ke²

Where;

k is the spring constant.

e is the extension.

Substituting into the formula, we have;

Workdone = ½ * 250 * 0.4²

Workdone = 125 * 0.16

Workdone = 20 Joules

5 0

2 years ago