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4 years ago

An object accelerates at 10 m/s2. If its starting velocity is 20 m/s and it travels for 30 seconds, what is its final velocity?

Physics

1 answer:

polet [3.4K]4 years ago

8 0

Answer:

320m/s

Explanation:

Step 1: State your values

s= NOT NEEDED

u=20m/s

v=?

a=10m/s²

t=30 seconds

Step 2: Find a suitable suvat equation:

v=u+at

Step 3:

subsitute values into equation:

?=20+(10x30)

20+300

320m/s

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An adiabatic closed system is accelerated from 0 m/s to 34 m/s. Determine the specific energy change of this system, in kJ/kg.

prohojiy [21]

Answer:

Δe=0.578 kJ/kg

Explanation:

Given data

Velocity v₁=0 m/s

Velocity v₂=34 m/s

to find

Specific energy change Δe

Solution

The specific energy change is simply determined from change in velocity

Δe=(v₂²-v₁²)/2

Put the given values to find the specific energy change

$=(\frac{(34)^{2} *10^{-3} }{2} )\\=0.578kJ/kg$

Δe=0.578 kJ/kg

6 0

3 years ago

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A baseball sits motionless near first base on a baseball diamond. What statement best explains why the baseball remains motionle

lukranit [14]

Answer:

Explanation:

Newton's first law states that, if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force. This postulate is known as the law of inertia.

hope this helps :))

5 0

3 years ago

A) A real object 4.0 cm high stands 30.0 cm in front of a converging lens of focal length 23 cm. Find the image distance, the im

vfiekz [6]

Explanation:

Given that,

Height of object = 4.0 cm

Distance of the object u= -30.0 cm

Focal length = 23 cm

We need to calculate the image distance

Using lens's formula

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Where, u = object distance

v = image distance

f = focal length

Put the value into the formula

$\dfrac{1}{v}-\dfrac{1}{-30}=\dfrac{1}{23}$

$\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{30}$

$\dfrac{1}{v}=\dfrac{7}{690}$

$v=98.57\ cm$

We need to calculate the height of the image

Using formula of height

$\dfrac{h'}{h}=\dfrac{-v}{u}$

Where, h' = height of image

h = height of object

$\dfrac{h'}{4}=\dfrac{-98.57}{-30}$

$h'=\dfrac{98.57\times4}{30}$

$h'=13.14\ cm$

The image is real and inverted.

(b). Now, object distance u = 13.0 cm

We need to calculate the image distance

Using lens's formula

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{13.0}$

$\dfrac{1}{v}=-\dfrac{10}{299}$

$v=-29.9\ cm$

We need to calculate the height of the image

$\dfrac{h'}{4}=\dfrac{-(-29.9)}{-13.0}$

$h=-\dfrac{29.9\times4}{13.0}$

$h=-9.2\ cm$

The image is virtual and erect.

Hence, This is required solution.

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4 years ago

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aleksandrvk [35]

Yes everyone knows this

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3 years ago

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Nine tree lights are connected inparallel across 120-V potential difference. The cord to the wall socket carries a current of 0.

jok3333 [9.3K]

Answer:

a)3000ohm

b)4.44mA

Explanation:

a) we were given a Nine tree lights connected inparallel across 120-V potential difference, since the resistor are in parallel we use the expresion below

1/R(total)= 1/R₁ + 1/R₂ + 1/R₃ + 1/R₃ +.... 1/R₉

But according to ohm'law which can be expressed below

V=IR

R=V/I

R(total)= 120/0.36

= 333.33ohm

1/R(total)= 1/R₁ + 1/R₂ + 1/R₃ + 1/R₃ +.... 1/R₉

R₁=R₂ =R₃ =R₄= R₅=R₆=R₇=R₈=R₉

1/R(total)=9/R

1/333.33= 9/R

R= 3000ohm

Therefore, the resistance is 3000ohm

b)the bulbs were connected in series here, then for series connection we use below expression

R₁=R₂ =R₃ =R₄= R₅=R₆=R₇=R₈=R₉

R(total)=9R

= 9*3000

=27000ohm

I=VR

I=V/R

I= 120/27000

= 4.44*10⁻³A

4.44mA

Therefore, the current is 4.4mA

7 0

4 years ago