All categories

3 years ago

An 1800-kg car has an acceleration of 3.8 m/s2. What is the force acting on the car?

Physics

2 answers:

Elina [12.6K]3 years ago

4 0

Answer:

F = ma

F = (1800 kg)(3.8 m/s2 )

F = 6840 kg m/s2 = 6800 N

Scorpion4ik [409]3 years ago

4 0

given

mass = 1800 kg

acceleration = 3.8m/s.s

force = m × a

= 1800 × 3.8

= 6840 N

You might be interested in

Pls say this a doubt i have

hodyreva [135]

Question four bulbs A,B,C and D are connected in a circuit shown in the figure below, the letters X, Y and Z represent three switches. Which switch is used to operate switch A separately?
Answer: x

4 0

3 years ago

A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.97-g pellet

dimaraw [331]

Answer:

$v = 2.8898 \frac{m}{s}$

Explanation:

This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.

When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.

<h3>Elastic potential energy</h3>

We know that a spring following Hooke's Law has a elastic potential energy:

$E_{ep} = \frac{1}{2} k (\Delta x)^2$

where $\Delta x$ is the displacement from the relaxed length and k is the spring's constant.

To obtain the spring's constant, we know that Hooke's law states that the force made by the spring is :

$\vec{F} = - k \Delta \vec{x}$

as we need 9.12 N to compress 4.60 cm, this means:

$k = \frac{9.12 \ N}{4.6 \ 10^{-2} \ m}$

$k = 198.26 \ \frac{ N}{m}$

So, the elastic energy of the compressed spring is:

$E_{ep} = \frac{1}{2} 198.26 \ \frac{ N}{m} (4.6 \ 10^{-2} \ m)^2$

$E_{ep} = 0.209759 \ Joules$

And when the spring is relaxed, the elastic potential energy will be zero.

<h3>Gravitational potential energy</h3>

To see how much gravitational potential energy will the pellet win, we can use

$\Delta E_{gp} = m g \Delta h$

where m is the mass of the pellet, g is the acceleration due to gravity and \Delta h is the difference in height.

Taking all this together, the gravitational potential energy when the spring is relaxed will be:

$\Delta E_{gp} = 4.97 \ 10^{-3} kg \ 9.8 \frac{m}{s^2} 4.6 \ 10^{-2} m$

$\Delta E_{gp} = 0.00224 \ Joules$

<h3>Kinetic Energy</h3>

We know that the kinetic energy for a mass m moving at speed v is:

$E_k = \frac{1}{2} m v^2$

so, for the pellet will be

$E_k = \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

<h3>All together</h3>

By conservation of energy, we know:

$E_{ep} = \Delta E_{gp} + E_k$

$0.209759 \ Joules = 0.00224 \ Joules + \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.209759 \ Joules - 0.00224 \ Joules$

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.207519 \ Joules$

$v = \sqrt{ \frac{ 0.207519 \ Joules}{ \frac{1}{2} \ 4.97 \ 10^{-3} kg } }$

$v = 2.8898 \frac{m}{s}$

7 0

3 years ago

An 11.0 -W energy-efficient fluorescent lightbulb is designed to produce the same illumination as a conventional 40.0-W incandes

yarga [219]

The amount or cost that the user of the energy-efficient bulb save during 100h of use will be $0.319.

<h3>How to calculate the cost?</h3>

For the 11.0W bulb, it should be noted that the value will be:

= 11.0 × 100 × (1/1000) × 0.110

= $0.121

The 40W bulb will be:

= 40 × 100 × (1/1000) × 0.110

= $0.44

Therefore, the amount that will be saved will be:

= $0.44 - $0.121

= $0.319

Learn more about cost on:

brainly.com/question/25109150

#SPJ4

6 0

2 years ago

Your backpack has a mass of 8 kg. You drop it from a height of 1.3m. How much work is done by gravity as the backpack falls?

olga_2 [115]

Answer:

The answer is C.

Explanation:

I guessed and it was right

6 0

3 years ago

Read 2 more answers

The work function for magnesium is 3.70 ev. what is its cutoff frequency?

alexandr402 [8]

The cutoff frequency for magnesium is 8.93 x 10¹⁴ Hz.

<h3>What is cutoff frequency?</h3>

The work function is related to the frequency as

W0 = h x fo

where, fo = cutoff frequency and h is the Planck's constant

Given is the work function for magnesium is 3.70 eV.

fo = 3.7 x 1.6 x 10⁻¹⁹ / 6.626 x 10⁻³⁴

fo = 8.93 x 10¹⁴ Hz.

Thus, the cut off frequency is 8.93 x 10¹⁴ Hz.

Learn more about cutoff frequency.

brainly.com/question/14378802

#SPJ1

7 0

2 years ago