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3 years ago

A car is traveling at 10 m/s. 10 seconds later the car is traveling 40 m/s. What is the car’s acceleration?

Physics

1 answer:

ratelena [41]3 years ago

3 0

Answer:

a = 3 m/s^2

Explanation:

Vi = 10 m/s

Vf = 40 m/s

t = 10 s

Plug those values into the following equation:

Vf = Vi + at

40 = 10 + 10a

---> a = 3 m/s^2

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If the radius of an atom is 60 pm and the radius of the Earth is 6000 km, by how many orders of

kherson [118]

Answer:

1 x 10¹⁷

Explanation:

Given data:

Radius of the earth = 6000km

Radius of an atom = 60pm

Now, how many orders is the radius of the earth larger than an atom

Solution:

To solve this problem, let us express both quantity as the same unit;

1000m = 1km

6000km = 6000 x 10³m = 6 x 10⁶m

60pm;

1 x 10⁻¹²m = 1pm

60pm = 60 x 1 x 10⁻¹²m = 6 x 10⁻¹¹m

Now;

The order: $\frac{6 x 10^{6} }{6 x 10^{-11} }$ = 1 x 10¹⁷

6 0

2 years ago

Select the correct answer<br>What is the importance of writing a draft?

givi [52]

Answer:to revise or edit

anything that can be made in the non draft one

Explanation:

4 0

3 years ago

A steady electric current flows through a wire. If 9.0 C of charge passes a particular spot in the wire in a time period of 2.0

defon

1) Current: 4.5 A

2) Time taken: 4.7 s

Explanation:

The electric current intensity is defined as the rate at which charge flows in a conductor; mathematically:

$I=\frac{q}{t}$

where

I is the current

q is the amount of charge passing a given point in a time t

For the wire in this problem, we have

q = 9.0 C is the amount of charge

t = 2.0 s is the time interval

Solving for I, we find the current:

$I=\frac{9.0}{2.0}=4.5 A$

To solve this problem, we can use again the same formula

$I=\frac{q}{t}$

where

I is the current

q is the amount of charge passing a given point in a time t

In this problem, we have:

I = 3.0 A (current)

q = 14.0 C (charge)

Therefore, the time taken for the charge to move past a particular spot in the wire is

$t=\frac{q}{I}=\frac{14.0}{3.0}=4.7 s$

Learn more about electric current:

brainly.com/question/4438943

brainly.com/question/10597501

brainly.com/question/12246020

#LearnwithBrainly

8 0

3 years ago

amping equipment weighing 6.0 kN is pulled across a frozen lake by means of ahorizontal rope. The coecient of kinetic friction i

Harman [31]

Answer:

Work done = 422.45 kJ

Explanation:

given,

weight of equipment = 6 kN

coefficient of kinetic friction = 0.05

distance up to which it is pulled = 1000 m

constant acceleration = 0.2 m/s²

Work done by the camper = ?

actual acceleration acting a'

m a = m a' - μ mg

a' = a + μ g

a' = 0.2 + 0.05 x 9.8

a' = 0.69 m/s²

Work done = Force x distance

F = m a'

$F = \dfrac{6000}{9.8} \times 0.69$

F = 422.44897 N

Work done = F x d

Work done = 422.44897 x 1000

Work done = 422449 J

Work done = 422.45 kJ

4 0

3 years ago

Lewiston and Vernonville are 208 miles apart. A car leaves Lewiston traveling towards Vernonville, and another car leaves Verno

iragen [17]

Answer:

Average speed of the car A = 70 miles per hour

Average speed of the car B = 60 miles per hour

Explanation:

Average speed of the car A is $v_{A} =\frac{x_{A} }{t_{A} }$ (Equation A) and Average speed of the car B is $v_{B} =\frac{x_{B} }{t_{B} }$ (Equation B), where $x_{A}$ and $x_{B}$ are the distances and $t_{A}$ and $t_{B}$ are the times at which are travelling the cars A and B respectively.