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Simora [160]
3 years ago
6

While waiting for his Mom to come out of the hairdresser's, Sean accidentally puts the car in gear and it begins to roll forward

. How far would the vehicle travel if it moved at 34 m/s for 2.5 s?
Physics
1 answer:
kicyunya [14]3 years ago
7 0

The distance travelled by the car is 85 m

Explanation:

The car in this problem here is travelling by uniform motion (so, at constant velocity and constant speed). Therefore, the distance covered by the car after a time t is given by:

d=vt

where

d is the distance covered

v is the speed

t is the time

For the car in this problem we have

v = 34 m/s

t = 2.5 s

Substituting into the formula, we find

d=(34)(2.5)=85 m

So, the car has travelled for 85 m.

Learn more about speed here:

brainly.com/question/8893949

#LearnwithBrainly

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Calculate the total mechanical energy of a 2 kg duck flying at 5 m/s, at a height of 10 meters above the ground.
dolphi86 [110]

Answer:

Total mechanical energy = 225 J

Explanation:

Given:

Mass of duck (m) = 2 kg

Speed of duck (v)= 5 m/s

Height of duck from ground (h) = 10 m

Gravitation acceleration (g) = 10 m/s²

Find:

Total mechanical energy

Computation:

Total mechanical energy = Kinetic energy + Potential energy

Total mechanical energy = (1/2)mv² + mgh

Total mechanical energy = (1/2)(2)(5)² + (2)(10)(10)

Total mechanical energy = 25 + 200

Total mechanical energy = 225 J

5 0
3 years ago
What are the stages of stellar evolution
Rudik [331]

Giant Molecular Cloud


Protostar


T-Tauri


Main Sequence


Subgiant, Red Giant, Supergiant


Core fusion


Red Giant, Supergiant


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Remnant

3 0
4 years ago
Read 2 more answers
A large truck is crossing a bridge 1 mile long. The bridge can only hold 14000 lbs, which is the exact weight of the truck. The
klasskru [66]
Depends on the weight of the bird.
1) Half a mile is about 800m.
2) 14000 lbs = about 6 tones, same like couple 4WD
3) Fuel consumption is about 20L per 100km or 0.2l each 1km or 0.16L within 800m
4) density of fuel is about 70% of density of water so .... weight of 1.6L of fuel burned would be about 1.6*0.7=1.1 kg

So if birds mass would be below 1 kg - the bridge will not collapse. But if it would be a pelican with the mass of 9kg - it would be a drama :)


  
8 0
3 years ago
Read 2 more answers
A uniform line charge of density λ lies on the x axis between x = 0 and x = L. Its total charge is 7 nC. The electric field at x
DedPeter [7]

Answer:

The electric field at x = 3L is 166.67 N/C

Solution:

As per the question:

The uniform line charge density on the x-axis for x, 0< x< L is \lambda

Total charge, Q = 7 nC = 7\times 10^{- 9} C

At x = 2L,

Electric field, \vec{E_{2L}} = 500N/C

Coulomb constant, K = 8.99\times 10^{9} N.m^{2}/C^{2}

Now, we know that:

\vec{E} = K\frac{Q}{x^{2}}

Also the line charge density:

\lambda = \frac{Q}{L}

Thus

Q = \lambda L

Now, for small element:

d\vec{E} = K\frac{dq}{x^{2}}

d\vec{E} = K\frac{\lambda }{x^{2}}dx

Integrating both the sides from x = L to x = 2L

\int_{0}^{E}d\vec{E_{2L}} = K\lambda \int_{L}^{2L}\frac{1}{x^{2}}dx

\vec{E_{2L}} = K\lambda[\frac{- 1}{x}]_{L}^{2L}] = K\frac{Q}{L}[frac{1}{2L}]

\vec{E_{2L}} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{2L}] = \frac{63}{L^{2}}

Similarly,

For the field in between the range 2L< x < 3L:

\int_{0}^{E}d\vec{E} = K\lambda \int_{2L}^{3L}\frac{1}{x^{2}}dx

\vec{E} = K\lambda[\frac{- 1}{x}]_{2L}^{3L}] = K\frac{Q}{L}[frac{1}{6L}]

\vec{E} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{6L}] = \frac{63}{6L^{2}}

Now,

If at x = 2L,

\vec{E_{2L}} = 500 N/C

Then at x = 3L:

\frac{\vec{E_{2L}}}{3} = \frac{500}{3} = 166.67 N/C

4 0
4 years ago
Can someone please help me with science
bixtya [17]

Answer:

Gravity as well as electrostatic and magnetic attraction and repulsion provide real life examples of forces being exerted by one object on another without them being in contact with each other.

Explanation:

Hope that helps!:)

8 0
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