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3 years ago

Part 1

Physics

1 answer:

nlexa [21]3 years ago

3 0

The force exerted by the hammer claws on the nail is =1578N

<h3 /><h3>How to calculate magnitude of force:</h3>

The force that is applied, as the effort on the hammer= F1

The force exerted by the hammer claws on the nail= F2

But,

magnitude of force exerted on hammer= F1*d1

While magnitude of force exerted by the hammer claws on the nail = F2( d2sin30°)

And, F1*d1 = F2( d2sin30°)

Where F1 = 133N

d1 = 29cm = 0.29m

F2 = X

d2 = 4.89cm = 0.0489m

After substitution,

133 × 0.29 = X( 0.0489 × sin30°)

38.57 = X ( 0.0489 × 0.5)

38.57 = X(0.02445)

X= 38.57/0.02445

X = 1578N

Therefore, the force exerted by the hammer claws on the nail is =1578N

Learn more about force here:

brainly.com/question/25239010

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Which of the following terms describes one’s ability to understand how another person is feeling?

liq [111]

Answer:

Empathy

Explanation:

Empathy: to understand the thoughts, feelings, or emotional state of another person

Appreciation: to show gratitude

Cooperation: is the association for mutual benefit

Commitment: is the agreement to do something.

From the definitions above, empathy is the term that describes one’s ability to understand how another person is feeling.

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3 years ago

What defines minerals will mark brainlest if croccted right

zvonat [6]

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3 years ago

A wooden block of mass M resting on a frictionless, horizontal surface is attached to a rigid rod of length ℓ and of negligible

mr Goodwill [35]

Answer:

$mv l$

$\frac{M }{(M + m)}$

Explanation:

Complete question statement is as follows :

A wooden block of mass M resting on a friction less, horizontal surface is attached to a rigid rod of length ℓ and of negligible mass. The rod is pivoted at the other end. A bullet of mass m traveling parallel to the horizontal surface and perpendicular to the rod with speed v hits the block and becomes embedded in it.

(a) What is the angular momentum of the bullet–block system about a vertical axis through the pivot? (Use any variable or symbol stated above as necessary.)

(b) What fraction of the original kinetic energy of the bullet is converted into internal energy in the system during the collision? (Use any variable or symbol stated above as necessary.)

$m$ = mass of the bullet

$v$ = velocity of the bullet before collision

$r$ = distance of the line of motion of bullet from pivot = $l$

$L$ = Angular momentum of the bullet-block system

Angular momentum of the bullet-block system is given as

$L = m v r$

$L = mv l$

$V$ = final velocity of bullet block combination

Using conservation of momentum

Angular momentum of bullet block combination = Angular momentum of bullet

$(M + m) V l = m v l\\V =\frac{mv}{(M + m)}$

$K_{o}$ = Initial kinetic energy of the bullet

Initial kinetic energy of the bullet is given as

$K_{o} = (0.5) m v^{2}$

$K_{f}$ = Final kinetic energy of bullet block combination

Final kinetic energy of bullet block combination is given as

$K_{f} = (0.5) (M + m) V^{2}$

Fraction of original kinetic energylost is given as

Fraction = $\frac{(K_{o} - K_{f})}{K_{o}} = \frac{((0.5) m v^{2} - (0.5) (M + m) V^{2})}{(0.5) m v^{2}}$

Fraction = $\frac{(m v^{2} - (M + m) (\frac{mv}{(M + m)})^{2})}{m v^{2}} = \frac{(Mm v^{2} + m^{2} v^{2} - m^{2} v^{2})}{(M + m) m v^{2}}$

Fraction = $\frac{(Mm v^{2} + m^{2} v^{2} - m^{2} v^{2})}{(M + m) m v^{2}}\\ \frac{M }{(M + m)}$

6 0

3 years ago

Increase the mass of the beanbag its gravitational potential energy will

kotegsom [21]

The gravitational potential energy will increase

Explanation:

The gravitational potential energy (GPE) of an object is the energy possessed by the object due to its position in a gravitational field.

Near the Earth's surface, the GPE of an object is given by

$GPE=mgh$

where

m is the mass of the object

g is the acceleration of gravity

h is the heigth of the object above the ground

From the equation, we see that the GPE is directly proportional to the mass: therefore, if the mass increases, the GPE will increase as well.

So, for the beanbag in this problem, when its mass increases, the GPE will increase as well.

Learn more about gravitational potential energy:

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#LearnwithBrainly

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3 years ago

If the potential due to a point charge is 490 V at a distance of 10 m, what are the sign and magnitude of the charge?

uysha [10]

Answer:

+5.4×10⁻⁷ C

Explanation:

Electric potential: This can be defined as the work done in bringing a unit charge from infinity to that point against the action of the field. The S.I unit of potential is volt (V)

The formula for potential is

V = kq/r............................ Equation 1

Where V = electric potential, k = proportionality constant, q = charge, r = distance.

making q the subject of the equation,

q = Vr/k............................ Equation 2

Given: V = 490 V, r = 10 m,

Constant: k = 9×10⁹ Nm²/C²

Substitute into equation 2

q = 490(10)/(9×10⁹)

q = 5.4×10⁻⁷ C

q = +5.4×10⁻⁷ C

Hence the charge is +5.4×10⁻⁷ C

7 0

3 years ago