All categories

4 years ago

An object accelerates from rest to a velocity of 4m/s over a distance of 20m what is the acceleration

Physics

1 answer:

Goshia [24]4 years ago

4 0

Answer:

Data:-vi=0 ,vf=4m/s ,s=20m ,a=?

Explanation:

Solution according to 3rd eq of motion 2as=vf²-vi² here we have to find a so a=vf²-vi²/2s ,a=(4²) -(0)²/2×20 ,a=16/40 ,a=0.4m/sec²

You might be interested in

What is a core sample ????

kap26 [50]

Answer:

Core samples are small portions of a formation taken from an existing well and used for geologic analysis. The sample is analyzed to determine porosity, permeability, fluid content, geologic age, and probable productivity of oil from the site. ... Core samples reveal the physical and chemical nature of the rock.

6 0

3 years ago

Who submits the federal budget every year?

artcher [175]

C) The president submits the federal budget every year.
Hope this helps you!

8 0

3 years ago

If the car is traveling at a velocity of 15 m/s, what is the approximate centripetal acceleration of the car?.

BaLLatris [955]

The vehicle's centripetal acceleration is equal to 22.5m/s²

Radius, r = 10 meter

Speed, V = 15 m/s

To ascertain the car's centripetal acceleration

A(c) = V²/R

We obtain the following when we enter the formula's parameters:

A(c) = 152/10

A(c) = 225/10

A(c) = 22.5m/s²

<h3>What is Centripetal acceleration ?</h3>

When an item moves in a circular route, one of its motion characteristics is centripetal acceleration. Any motion in a circle with an acceleration vector pointing in the direction of the circle's centre is referred to as centripetal acceleration.

Centripetal forces cause accelerations at the centripetal axis. With the exception of the Earth's rotation around the Sun, any satellite's circular motion around a celestial body is brought on by the centripetal force produced by their mutual gravitational pull.

Hence, Centripetal acceleration is

22.5 m/s²

Learn more about Centripetal acceleration here:

brainly.com/question/79801

#SPJ4

7 0

1 year ago

classroom and draw an approxim object Three forces are acting on an object (Figure 1.32) which is in equilibrium. Determine forc

kumpel [21]

I’m not good with physics but I’m good with the theory what goes up must come down

6 0

3 years ago

I need homework help

KIM [24]

1) the weight of an object at Earth's surface is given by $F=mg$ , where m is the mass of the object and $g=9.81 m/s^2$ is the gravitational acceleration at Earth's surface. The book in this problem has a mass of m=2.2 kg, therefore its weight is
$F=mg=(2.2 kg)(9.81 m/s^2)=21.6 N$

2) On Mars, the value of the gravitational acceleration is different: $g=3.7 m/s^2$ . The formula to calculate the weight of the object on Mars is still the same, but we have to use this value of g instead of the one on Earth: $F=mg=(2.2 kg)(3.7 m/s^2)=8.1 N$

3) The weight of the textbook on Venus is F=19.6 N. We already know its mass (m=2.2 kg), therefore by re-arranging the usual equation F=mg, we can find the value of the gravitational acceleration g on Venus:
$g= \frac{F}{m}= \frac{ 19.6 N}{2.2 kg}=8.9 m/s^2$

4) The mass of the pair of running shoes is m=0.5 kg. Their weight is F=11.55 N, therefore we can find the value of the gravitational acceleration g on Jupiter by re-arranging the usual equation F=mg:
$g= \frac{F}{m} = \frac{11.55 N}{0.5 kg} =23.1 m/s^2$

5) The weight of the pair of shoes of m=0.5 kg on Pluto is F=0.3 N. As in the previous step, we can calculate the strength of the gravity g on Pluto as
$g= \frac{F}{m} = \frac{0.3 N}{0.5 kg} =0.6 m/s^2$

<span>6) On Earth, the gravity acceleration is </span> $g=9.81 m/s^2$ <span>. The mass of the pair of shoes is m=0.5 kg, therefore their weight on Earth is
</span> $F=mg=(0.5 kg)(9.81 m/s^2)=4.9 N$ <span>
</span>

5 0

3 years ago