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Fittoniya [83]
1 year ago
10

How long did it take Fernando to drive to

Physics
1 answer:
barxatty [35]1 year ago
3 0
How do you calculate distance over speed?
Image result for if your going 30 m/s to go somewhere thats 1680 miles away
The formula can be rearranged in three ways:
speed = distance ÷ time.
distance = speed × time.
time = distance ÷ speed.

1680÷30 = 56

So it would take around 56 minutes to get to Kroger.


I hope this helps !! :)
You might be interested in
A current of 0. 82 a flows through a light bulb. how much charge passes through the light bulb during 94 s?
gladu [14]

A current of 0. 82A flows through a light bulb. The charge passed through the light bulb during 94 s is 77.08C

The amount of charge flown for a given period of time determines the current passed through a bulb or electrical body.

The relation between the charge, current and time is given as:

Q = I × t

where, Q is the charge flown through bulb

I is the current passed through bulb

t is the time for which charge passes through bulb

Given,

I = 0.82A

t = 94s

Q = ?

Substituting the values in the above formula:

Q = I × t

Q = 0.82 × 94

Q = 77.08C

Hence, The charge passed through the light bulb during 94 s is 77.08C

Learn more about Current here, brainly.com/question/2264542

#SPJ4

4 0
1 year ago
Fiora starts riding her bike at 20 mi/h. after a while, she slows down to 12 mi/h, and maintains that speed for the rest of the
hammer [34]
<span>d = r*t

t = hours at 20 mi/hr


20t + 12*(4.5 - t) = 70
8t = 16
t = 2 hours

d at 20 mi/hr = 20*2 = 40 miles

40/20 + 30/12 = 4.5 hours

Fiora travels a total distance of 4.5 hours</span>
3 0
3 years ago
Read 2 more answers
A puck of mass 0.70 kg approaches a second, identical puck that is stationary on frictionless ice. The initial speed of the movi
natali 33 [55]

Answer:

  • v_1  =  \ 5.196 \frac{m}{s}
  • v_2 =  3 \frac{m}{s}

Explanation:

For this problem, we just need to remember conservation of momentum, as there are no external forces in the horizontal direction:

\vec{p}_i = \vec{p}_f

where the suffix i  means initial, and the suffix f means final.

The initial momentum will be:

\vec{p}_i = m_1 \ \vec{v}_{1_i} + m_2 \ \vec{v}_{2_i}

as the second puck is initially at rest:

\vec{v}_{2_i} = 0

Using the unit vector \vec{i} pointing in the original line of motion:

\vec{v}_{1_i} = 6.0 \frac{m}{s} \hat{i}

\vec{p}_i = 0.70 \ kg  \ 6.0 \frac{m}{s} \ \hat{i} + 0.70 \ kg \ 0

\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i}

So:

\vec{p}_i =  4.2 \ \frac{kg \ m}{s} \ \hat{i} = \vec{p}_f

\vec{p}_f =  4.2 \ \frac{kg \ m}{s} \ \hat{i}

Knowing the magnitude and directions relative to the x axis, we can find Cartesian representation of the vectors using the formula

\ \vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )

So, our velocity vectors will be:

\vec{v}_{1_f} = v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ )

\vec{v}_{2_f} = v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )

We got

\vec{p}_f = 0.7 \ kg \ \vec{v}_{1_f} + 0.7 \ kg \ \vec{v}_{2_f}

4.2 \ \frac{kg \ m}{s} \ \hat{i} = 0.7 \ kg \   v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ )  + 0.7 \ kg \ v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )

So, we got the equations:

4.2 \ \frac{kg \ m}{s}  = 0.7 \ kg \   v_1 \  cos(30 \°) + 0.7 \ kg \ v_2 \  cos(-60 \°)

and

0  = 0.7 \ kg \   v_1 \  sin(30 \°) + 0.7 \ kg \ v_2 \  sin(-60 \°).

From the last one, we get:

0  = 0.7 \ kg \  ( v_1 \  sin(30 \°) +  \ v_2 \  sin(-60 \°) )

0  =  v_1 \  sin(30 \°) +  \ v_2 \  sin(-60 \°)

v_1 \  sin(30 \°) = -  \ v_2 \  sin(-60 \°)

v_1  =  \ v_2 \  \frac{sin(60 \°)}{ sin(30 \°) }

and, for the first one:

4.2 \ \frac{kg \ m}{s}  = 0.7 \ kg  \ (  v_1 \  cos(30 \°) + v_2 \  cos(60 \°) )

\frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} =    v_1 \  cos(30 \°) + v_2 \  cos(60 \°)

\frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} =    v_1 \  cos(30 \°) + v_2 \  cos(60 \°)

6 \ \frac{m}{s} =    (\ v_2 \  \frac{sin(60 \°)}{ sin(30 \°) } ) \  cos(30 \°) + v_2 \  cos(60 \°)

6 \ \frac{m}{s} = v_2     (\   \frac{sin(60 \°)}{ sin(30 \°) } ) \  cos(30 \°) +   cos(60 \°)

6 \ \frac{m}{s} = v_2  * 2

so:

v_2 = 6 \ \frac{m}{s} / 2 = 3 \frac{m}{s}

and

v_1  =  \ 3 \frac{m}{s}  \  \frac{sin(60 \°)}{ sin(30 \°) }

v_1  =  \ 5.196 \frac{m}{s}

3 0
3 years ago
Suppose the entire solar nebula had cooled to 50 K before the solar wind cleared the early solar system of its gases. How would
Butoxors [25]

Suppose the entire solar nebula had cooled to 50 K before the solar wind cleared the early solar system of its gases. How would the composition and sizes of the planets of the inner solar system be different from what we see today is given below

Explanation:

1.In astronomy or planetary science, the frost line, also known as the snow line or ice line, is the particular distance in the solar nebula from the central protostar where it is cold enough for volatile compounds such as water, ammonia, methane, carbon dioxide, carbon monoxide to condense into solid ice grains.

2.The frost line in the solar nebula lies between Mars and Jupiter. It is the distance where it was cold enough for hydrogen compounds to condense into ices. Frost line: Explain how temperature differences led to the formation of two distinct types of planets.

3. The frost line is the point moving away from the Sun where it is cool enough for hydrogen compounds to freeze. Since the solar nebula was hotter near the center of the disk, hydrogen compounds such as water stayed gaseous in the inner solar system. Outside of the frost line, they froze.

4.The solar nebula flattened into a rotating disk. As gas became dense and hot, then it spins faster and pulled towards the center whereby the sun is formed. Solar nebula they collapse where the protostellar disk rotates. In the center of of nebula, there is a fusion begins and then sun is being formed

5.When it comes to the formation of our Solar System, the most widely accepted view is known as the Nebular Hypothesis. In essence, this theory states that the Sun, the planets, and all other objects in the Solar System formed from nebulous material billions of years ago.

4 0
3 years ago
An airplane which intends to fly due south at 250 km/hr experiences a wind blowing westward at 40 km/hr. What is the actual spee
sleet_krkn [62]

Answer:

simple is rumple a daily ok I'll be

7 0
2 years ago
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