If a car's mass is 439 kg, what is its weight on earth?

You are driving to the grocery store at 20 m/s. You are 110 m from an intersection when the light turns red. You have a reaction

Lynna [10]

Answer:

a)

reaction time = 0.70 s

distance travelled in reaction time = v*t

= 20 m/s * 0.70 s

= 14 m

So, when brake is applied, distance remaining= 110 m - 14 m = 96 m

Answer: 96 m

b)

vf = 0 m/s

d = 96 m

vi = 20 m/s

use:

vf^2 = vi^2 + 2*a*d

0 = 20^2 + 2*a*96

-400 = 2*a*96

a = -2.08 m/s^2

Answer: -2.08 m/s^2

c)

use:

vf = vi + a*t

0 = 20 - 2.08*t

t = 9.6 s

Answer: 9.6 s

Explanation:

8 0

2 years ago

For how long a time must a tow truck pull with a force of 550N on a stalled 1200kg car to give it a forward velocity of 2.0m/s?

horsena [70]

Answer:

4.36 seconds

Explanation:

According to the question;

Force is 550 N
Mass of the car is 1200 kg
Velocity of the car is 2.0 m/s

We are needed to find the time the car must the tow track pull the car.

From Newton's second law of motion;
Impulsive force, F = Mv÷t , where m is the mass, v is the velocity and t is the time.

Rearranging the formula;

t = mv ÷ F

Thus;

Time = (1200 kg × 2.0 m/s²) ÷ 550 N

= 4.36 seconds

Thus, the time needed to pull the car is 4.36 seconds

5 0

3 years ago

After hitting the spring, the block is bounced back up the ramp. The maximum compression of the spring is Δx=0.03m, and the spri

Lyrx [107]

Answer:

v = 1.28 m/s

Explanation:

Given that,

Maximum compression of the spring, $\Delta x=0.03\ m$

Spring constant, k = 800 N/m

Mass of the block, m = 0.2 kg

To find,

The velocity of the block when it first reaches a height of 0.1 m above the ground on the ramp.

Solution,

When the block is bounced back up the ramp, the total energy of the system remains conserved. Let v is the velocity of the block such that,

Initial energy = Final energy

$\dfrac{1}{2}kx^2=mgh+\dfrac{1}{2}mv^2$

Substituting all the values in above equation,

$\dfrac{1}{2}\times 800\times 0.03^2=0.2\times 9.8\times 0.1+\dfrac{1}{2}\times 0.2\times v^2$

v = 1.28 m/s

Therefore the velocity of block when it first reaches a height of 0.1 m above the ground on the ramp is 1.28 m/s.

5 0

2 years ago

Which wavelength of light is the best choice when attempting to quantitatively relate solution absorbance and concentration?.

ANEK [815]

The optimum wavelength is 450 nm because that is the wavelength of maximum absorbance by FeSCN2+(aq)

you should choose a wavelength with maximum absorbance. In this case, you are using the scattered light, not the absorbed light as your signal. So you should avoid wavelengths where there are absorption peaks.

<h3>What is wavelength ?</h3>

A waveform signal that is carried in space or down a wire has a wavelength, which is the separation between two identical places (adjacent crests) in the consecutive cycles. This length is typically defined in wireless systems in metres (m), centimetres (cm), or millimetres (mm) (mm).

The distance between two waves' crests serves as an illustration of wavelength. When you and another person have the same overall mindset and can easily communicate, that is an example of being on the same wavelength.

Learn more about Wavelength here:

brainly.com/question/10750459

#SPJ4

6 0

2 years ago

A speed-time graph is shown below:

VladimirAG [237]

Answer: $0.5 m/s^{2}$

Explanation:

Average acceleration $a_{ave}$ is the variation of velocity $\Delta V$ over a specified period of time $\Delta t$ :

$a_{ave}=\frac{\Delta V}{\Delta t}}$

Where:

$\Delta V=V_{f}-V_{o}$ being $V_{o}=0 cm/s$ the initial velocity and $V_{f}=4 cm/s$ the final velocity (according to the information given from the described graph)

$\Delta t=8 s$

Then:

$a_{ave}=\frac{4 cm/s -0 cm/s}{8 s}}$

$a_{ave}=0.5 m/s^{2}$

5 0

3 years ago