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3 years ago

A center-seeking force related to acceleration is _______ force.

1 answer:

5 0

<span>The question is 'a centre seeking force related to acceleration is ............... force. The answer is centripetal force. Motion in a curved path is an accelerated motion and it requires a force that will direct the moving object towards the centre of curvature of the path of motion. This centre seeking force is known as centripetal force.</span>

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A changing magnetic field can produce an electric current. True or False?

Mamont248 [21]

The answer is True :) have a good day

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3 years ago

A basketball is held over head at a height of 2.4 m. The ball is lobbed to a teammate at 8 m/s at an angle of 40'. If the ball i

cupoosta [38]

Explanation:

since both the teammates are of the same height, their height won't matter. Because now the basketball won't cover any vertical distance.

We have to calculate its range the horizontal distance covered by it when tossed from one teammate to the other.

range can be calculated by the formula :-

$\boxed{\mathfrak{range = \frac{ u {}^{2} \sin 2\theta }{g} }}$

u is the velocity during its take off and $\theta$ is the angle at which its thrown

Given that

u = 8m/ s
$\theta$ = 40°

calculating range using the above formula

$= \frac{ {8}^{2} \sin2(40) }{10}$

$= \frac{64 \times \sin(80) }{10}$

value of sin 80 = 0. 985

$= \frac{64 \times 0.985}{10}$

$= \frac{63.027}{10}$

$= 6.3027$

Hence,

$\mathfrak { \blue{the \: teammate \: is \: \red{\underline{6.3027 \: meters} }\: away } }$

7 0

3 years ago

What do we call the center of an atom that contains the mass of the<br> with<br> ca

Delvig [45]

Answer:

Hope this helps! Please Mark Brainliest!

Explanation:

The Nucleus: The Center of an Atom. The nucleus, that dense central core of the atom, contains both protons and neutrons. Electrons are outside the nucleus in energy levels.

6 0

3 years ago

Carol has several test tubes of different liquids. If Carol removes heat from the liquid substances, which of the following is m

Archy [21]

Answer:

Explanation:

6 0

3 years ago

A 0.950 kg block is attached to a spring with spring constant 16.0 N/m . While the block is sitting at rest, a student hits it w

Bumek [7]

Answer:

Explanation:

Given that,

Mass attached m = 0.95kg

Spring constant k = 16N/m

Instantaneous speed v = 36cm/s = 0.36m/s

Amplitude A=?

When x = 0.7A

Using conservation of energy

∆K.E + ∆P.E = 0

K.E(final) — K.E(initial) + P.E(final) — P.E(initial) = 0

At the beginning immediately the hammer hits the mass, the potential energy is 0J, Therefore, P.E(initial) = 0J, so the speed is maximum.

Also, at the end, at maximum displacement, the speed is zero, therefore, K.E(final) = 0

So, the equation becomes

— K.E(initial) + P.E(final) = 0

K.E(initial) = P.E(final)

½mv² = ½kA²

mv² = kA²

0.95 × 0.36² = 16×A²

0.12312 = 16•A²

A² = 0.12312/16

A² = 0.007695

A = √0.007695

A = 0.088 m

A = 8.8cm

B. Speed at x = 0.7A

Using the same principle above

K.E(initial) = P.E(final)

½mv² = ½kA²

Where A = 0.7A = 0.7 × 0.088 = 0.0614m

Then,

½× 0.95 × v² = ½ × 16 × 0.0614²

0.475v² = 0.0310644

v² = 0.0310644/0.475

v² = 0.0635

v = √0.0635

v = 0.252 m/s

v = 25.2 cm/s

8 0

3 years ago