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bazaltina [42]
3 years ago
12

Please help me with this question​

Physics
1 answer:
Oksanka [162]3 years ago
6 0

Car at rest:

velocity= 0m/s

Acceleration:

0.2m/s²

Since total time:

3 min = 180s

Formula of acceleration:

acceleration = [final velocity - initial velocity] ÷ [total time]

Velocity at end:

0.2m/s² = [final velocity - 0m/s] ÷ [180s]

0.2m/s² × 180s = [final velocity]

[final velocity] = 36m/s

Distance travelled:

Velocity = displacement(distance) ÷ time

36m/s = displacement(distance) ÷ 180s

displacement(distance) = 36m/s × 180s

displacement(distance) = 6480m

<em><u>Hey I'm sorry but i do not understand why the answer on your worksheet for distance travelled is 3240m... its </u></em><em><u>half</u></em><em><u> of what my answer is...</u></em>

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What is the expermintal example of Zeeman effect?
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Answer:

When the spectral lines are absorption lines, the effect is called inverse Zeeman effect.

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3 years ago
A car drives over a hilltop that has a radius of curvature 0.120 km at the top of the hill. At what speed would the car be trave
Mashutka [201]

Answer:

34.3 m/s

Explanation:

Newton's Second Law states that the resultant of the forces acting on the car is equal to the product between the mass of the car, m, and the centripetal acceleration a_c (because the car is moving of circular motion). So at the top of the hill the equation of the forces is:

mg-R = m a_c = m\frac{v^2}{r}

where

(mg) is the weight of the car (downward), with m being the car's mass and g=9.8 m/s^2 is the acceleration due to gravity

R is the normal reaction exerted by the road on the car (upward, so with negative sign)

v is the speed of the car

r = 0.120 km = 120 m is the radius of the curve

The problem is asking for the speed that the car would have when it tires just barely lose contact with the road: this means requiring that the normal reaction is zero, R=0. Substituting into the equation and solving for v, we find:

v=\sqrt{gr}=\sqrt{(9.8 m/s^2)(120 m)}=34.3 m/s

6 0
3 years ago
A lever has a mechanical advantage of 5.4. If you apply a force of 500 N to the lever, how much force does the lever apply to th
Yanka [14]

Force applied by the lever to the object=2700 N

Explanation:

Mechanical advantage= \frac{LOAD}{EFFORT}

mechanical advantage= 5.4

effort force= 500 N

we need the load force

5.4= load/500

load=2700 N

Thus the force applied by the lever to the object=2700 N

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3 years ago
Protons in an atomic nucleus are typically 10−15 m apart. what is the electric force (in n) of repulsion between nuclear protons
dybincka [34]
<span>The electric force is given by: 
 F = [ k*(q1)*(q2) ] / d^2 
 F = Electric force 
 k = Coulomb's constant 
 q1 = Charge of one proton 
 q2 = Charge of second proton 
 d = Distance between centers of mass 
 Values: 
 F = unknown 
 k = 8.98E 9 N-m^2/C^2 
 q1 = 1.6E-19 
 q2 = 1.6E-19 
 d = 1.0E-15 m 
 Insert values into F = [ k*(q1)*(q2) ] / d^2 
 F = [ (8.98E 9 N-m^2/C^2) * (1.6E-19) * (1.6E-19) ] / (1.0E-15 m)^2 
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 answer
 the electric force of repulsion between nuclear protons is 229.888 N

3 0
3 years ago
Read 2 more answers
Perform the calculation and report your answer using sig figs. 657.70 - 26.543
Anton [14]

Answer:

The answer is 631.157

Explanation:

The question requested that the answer to the subtraction of 26.543 from 657.70 must be written using significant figures.

Here are a few tips about how to Identify significant figures.

1) It should be noted that <u>the number "0" is what is usually (but not always) affected</u> while trying to identify significant figures. Hence, <u>all other numbers/digits are always significant</u>. For example, 26.543 has five significant figures.

2) The zeros found between these "other numbers/digits" are also significant. For example, 2202 has four significant figures.

3) In the case of a decimal, the tailing zeros or the final zero is also significant. 657.70 and 657.07 have five significant figures.

Now, back to the question

657.70  - 26.543  = 631.157.

Our final answer does not have a zero, hence all the digits (six) are significant.

8 0
3 years ago
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