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3 years ago

Where are the magnetic field and force weakest in a magnet?

Physics

1 answer:

nadya68 [22]3 years ago

5 0

The magnetic field and force is weakest in the center of the magnet. The correct answer is C.

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Suppose you are an astronaut and you have been stationed on a distant planet. You would like to find the acceleration due to the

Maksim231197 [3]

Answer:

Explanation:

Using the equation of motion expressed as v = u+gt where;

v is the final velocity of the ball

u is the initial velocity

g is the acceleration due to gravity

t is the time taken

Given

u = 9m/s

v = 0m/s

t = 6.4s

Required

acceleration due to gravity g

Since the rock is thrown up, g will be a negative value.

v = u+(-g)t

0 = 9-6.4g

-9 = -6.4g

6.4g = 9

divide both sides by 6.4

6.4g/6.4 = 9/6.4

g = 1.40625m/s²

Hence the acceleration due to gravity on the planet is 1.40625m/s²

6 0

3 years ago

Which of the following types of electromagnetic radiation has the longest wavelength? choose one answer.

Setler79 [48]

The right answer is red light

8 0

2 years ago

How do you find circular motion?

MatroZZZ [7]

Answer:

Explanation:

By multiplying the rotational frequency with the circumference we can determine the average speed of the object. The circular velocity formula is expressed as, vc = 2 πr / T. Where in, r denotes the radius of the circular orbit. T is time period.

6 0

3 years ago

Read 2 more answers

When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less dam

Len [333]

Answer:

Explanation:

Given that,

Mass of the heavier car m_1 = 1750 kg

Mass of the lighter car m_2 = 1350 kg

The speed of the lighter car just after collision can be represented as follows

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2}$

$v_2=\frac{(1850)(1.4)+(1450)(-1.10)-(1850)(0.250)}{1450} \\\\=\frac{2590+(-1595)-(462.5)}{1450} \\\\=\frac{2590-1595-462.5}{1450} \\\\=\frac{532.5}{1450}\\\\=0.367m/s$

b) the change in the combined kinetic energy of the two-car system during this collision

$\Delta K.E=(\frac{1}{2} m_1v_1^2+\frac{1}{2} m_2v_2^2)-(\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u_2^2)\\\\=\frac{1}{2} (m_1(v_1^2-u_1^2)+m_2(v_2^2-u_2^2))$

substitute the value in the equation above

$=\frac{1}{2} (1850((0.250)^2-(1.4)^2)+(1450((0.3670)^2-(-1.10)^2)\\\\=\frac{1}{2}(11850(0.0625-1.96)+(1450(0.1347)-(1.21))\\\\= \frac{1}{2}(11850(-1.8975))+(1450(-1.0753))\\\\=\frac{1}{2} (-3510.375+(-1559.185)\\\\=\frac{1}{2} (-5069.56)\\\\=-2534.78J$

Hence, the change in combine kinetic energy is -2534.78J

8 0

3 years ago

Suppose you ran 53 km in 67 min. With what speed did you run?

melomori [17]

Answer:

47.46 kilometers an hour

Explanation:

hope this helped

4 0

2 years ago

Read 2 more answers