All categories

4 years ago

A Tesla Roadster car accelerates from rest at a rate of 7.1m/s for a time of 3.9s

Physics

1 answer:

liberstina [14]4 years ago

5 0

The distance covered is 54.0 m

Explanation:

Since the motion of the car is a uniformly accelerated motion, we can use the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where

u is the initial velocity

t is the time

a is the acceleration

s is the distance covered

For the car in this problem, we have

u = 0 (it starts from rest)

$a=7.1 m/s^2$ is the acceleration

t = 3.9 s is the time

Substituting, we find s:

$s=0+\frac{1}{2}(7.1)(3.9)^2=54.0 m$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

You might be interested in

Don't mind me I'm trying to get banned

Vilka [71]

Answer is True

Have a good day !

8 0

3 years ago

Please help. All the information is in the image.

kondaur [170]

The tiger's position at time t is given by

x (horizontal) = (4.5 m/s) t

y (vertical) = 7.5 m - 1/2 gt ²

Solve y = 0 for t to find the time it takes for the tiger to reach the ground :

0 = 7.5 m - 1/2 (9.8 m/s²) t ²

===> t = √(2 (7.5 m) / (9.8 m/s²)) ≈ 1.2 s

Evaluate x at this time :

x = (4.5 m/s) (1.2 s) ≈ 5.6 m

5 0

3 years ago

A block is at rest on the incline shown in the gure. The coefficients of static and kinetic friction are 0.6 and 0.51, respec- t

Alekssandra [29.7K]

Normal reaction force on the block while it is at rest on the inclined plane is given as

$F_n = mgcos\theta$

here we know that

m = 46 kg

$\theta = 29^o$

now we will have

$F_n = 46*9.8*cos29 = 394.3 N$

now the limiting friction or maximum value of static friction on the block will be given as

$F_s = \mu_s * F_n$

$F_s = 0.6 * 394.3 = 236.56 N$

Above value is the maximum value of force at which block will not slide

Now the weight of the block which is parallel to inclined plane is given as

$F_{||} = mg sin\theta$

here we know that

$F_{||} = 46*9.8 sin29 = 218.55 N$

Now since the weight of the block here is less than the value of limiting friction force and also the block is at rest then the frictional force on the block is static friction and it will just counter balance the weight of the block along the inclined plane.

So here friction force on the given block will be same as its component on weight which is 218.55 N

5 0

3 years ago

A block has a volume of 0.09 m3 and a density of 4,000 kg/m3. What's the force of gravity acting on the block in water?

NeX [460]

If one m³ of that material holds 4,000 kg of it,
then 0.09 m³ holds

(0.09) x (4,000) = 360 kg of it

The force of gravity acting on 360 kg of anything
on the Earth's surface is

(mass) x (gravity)

= (360 kg) x (9.8 m/s²) = 3,528

5 0

4 years ago

Suppose a 1 Gbps point-to-point link is being set up between the Earth and a new lunar colony. The distance from the moon to Ear

vagabundo [1.1K]

Answer:

The time required to send the data from Earth to Moon will be 1.28s while for a two way communication, to send it back to the earth, it will take double time i.e. RTT = 2.56s

Explanation:

Distance between Earth and Moon = 385,000 km = 3.85 x 10⁸m

Speed of data travel = speed of light ≈ 3 x 10⁸m/s

As, v=d/t

t=d/v

$t=\frac{3.85*10^{8} }{3*10^{8}}$

t=1.28s

RTT = Double of single way time taken = 2x1.28

RTT=2.56s

7 0

3 years ago