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patriot [66]
3 years ago
14

What is the average speed of a car traveling for 1000 miles in 6 hours

Physics
1 answer:
nlexa [21]3 years ago
8 0

You would divide 6 into 1000 so your answer is B. 166

You might be interested in
If the star Sirius emits 23 times more energy than the Sun, why does the Sun appear brighter in the sky?
Ganezh [65]

Answer:

As b ∝ (L/r²) and

the distance of the sun from the earth is 0.00001581 light years

and

the distance of the Sirius from the earth is 8.6 light years

hence,

the Sun appear brighter in the sky

Explanation:

The brightness (b) is directly proportional to the Luminosity of the star (L) and inversely proportional to the square of the distance between the star and the observer (r).

thus, mathematically,

b ∝ (L/r²)

now,

given

L for sirius is 23 times more than the sun i.e 23L

now,

the distance of the sun from the earth is 0.00001581 light years

and

the distance of the Sirius from the earth is 8.6 light years

thus,

using the the relation between conclude that the value of brightness for the Sirius comes very very low as compared to the value for brightness for the Sun.

hence, the sun appears brighter

5 0
3 years ago
A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before
Darina [25.2K]

{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}

\\

{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before  \: it \: comes \: to \: rest \:( s_{2} )}

\\

{\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity   = v\:m/s} \\\\

\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \:  s_{1} \: penetration =  \dfrac{v}{2}  \:m/s} \\\\

\:\:\:\:\bullet\:\:\:\sf{s_{1} =  \dfrac{3}{100}  = 0.03 \: m}

\\

☯ As we know that,

\\

\dashrightarrow\:\: \sf{ {v}^{2}  =  {u}^{2} + 2as }

\\

\dashrightarrow\:\: \sf{  \bigg(\dfrac{v}{2} \bigg)^{2}  =  {v}^{2}   + 2a s_{1}}

\\

\dashrightarrow\:\: \sf{  \dfrac{ {v}^{2} }{4}  =  {v}^{2}  + 2 \times a \times 0.03  }

\\

\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4}  -  {v}^{2}  = 0.06 \times a  }

\\

\dashrightarrow\:\: \sf{\dfrac{ -  3{v}^{2} }{4}  = 0.06 \times a  }

\\

\dashrightarrow\:\: \sf{a =  \dfrac{ - 3 {v}^{2} }{4 \times 0.06}  }

\\

\dashrightarrow\:\: \sf{ a =  \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }

\\

\:\:\:\:\bullet\:\:\:\sf{  Initial\:velocity=v\:m/s} \\\\

\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }

\\

\dashrightarrow\:\: \sf{  {v}^{2}  =  {u}^{2}  + 2as}

\\

\dashrightarrow\:\: \sf{{0}^{2}  =  {v}^{2}  + 2 \times  \dfrac{ - 25 {v}^{2} }{2}  \times s  }

\\

\dashrightarrow\:\: \sf{ -  {v}^{2}  =  - 25 {v}^{2}  \times s  }

\\

\dashrightarrow\:\: \sf{  s =  \dfrac{ -  {v}^{2} }{ - 25 {v}^{2} }}

\\

\dashrightarrow\:\: \sf{  s =  \dfrac{1}{25} }

\\

\dashrightarrow\:\: \sf{ s = 0.04 \: m }

\\

☯ For left penetration (s₂)

\\

\dashrightarrow\:\: \sf{s =  s_{1} +  s_{2}  }

\\

\dashrightarrow\:\: \sf{  0.04 = 0.03 +  s_{2}}

\\

\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }

\\

\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}

\\

\star\:\sf{Left \: penetration \: before  \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\

4 0
2 years ago
Why do we say that a baseball falls toward Earth, and not Earth toward the baseball?
mezya [45]
Probably because the earth is stationary rotating on its axis whereas the baseball is the one doing all of the movement after being thrown
4 0
3 years ago
Cassie is an engineering undergrad. She does not like talking much, and has become a quiet and withdrawn person while in her pro
stepan [7]

Answer:

A

Explanation:

Analytical people are less responsive to others. They hence tend to focus more on work than people and are less interested in leading, being happier to work by themselves. They may be prudent and systematic, making them good at analytic work.

Some character traits of Analytical people

Focuses on work and working more than people

Likes to be correct and will take time to ensure this

Thoughtful, careful fact-oriented and precise

Good at problem-solving

Likes organization and structure

Avoids working in a group, preferring to work alone

Can be over-critical and unresponsive

Cautious in decision-making

When stressed may withdraw or become headstrong

8 0
3 years ago
Narysuj wykres zależności v(t) jeśli w chwili początkowej t=0 V=10m/s w każdej sekundzie szybkość zmniejsza się o 1m/s . Po jaki
irina1246 [14]

1) See graph in attachment

2) 10 s

3) 50 m

Explanation:

1)

In this problem, we have an object initially moving with a velocity of

v = 10 m/s

when the time is

t = 0 s

Then, we are told that the speed of the object is decreasing by 1 m/s every  second. This means that on a velocity-time graph, the motion will be represented by a straight line, starting from v = 10 when t = 0, and decreasing by 1 m/s every second.

The result can be found in the graph in attachment.

Moreover, we can also infer that the motion of the object is accelerated (because velocity is changing), and that the acceleration is constant and it is equal to

a=1 m/s^2

which is equivalent to the gradient of the line in the velocity-time graph.

2)

In this part, we want to find after what time the body will stop its motion.

To do that, we can use the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

In this problem:

u = 10 m/s is the initial velocity of the body

a=-1 m/s^2 is the acceleration

v = 0 m/s, because we want to find the time T at which the body will stop

Re-arranging the equation, we find:

T=-\frac{u}{a}=-\frac{10}{-1}=10 s

3)

In order to find the total distance covered by the body during its accelerated motion, we have to use another suvat equation:

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

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

In this problem:

u = 10 m/s is the initial velocity

a=-1 m/s^2 is the acceleration

t = 10 s is the time it takes for the body to stop (found in part 2)

Solving for s, we find the distance covered:

s=(10)(10)+\frac{1}{2}(-1)(10)^2=50 m

7 0
3 years ago
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