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2 years ago

What is the mass of a bullet in free fall with a force of 300N?

Physics

1 answer:

Valentin [98]2 years ago

8 0

There’s not enough information

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There are ___________ of galaxies in the universe containing __________ of stars in each galaxy.

aliya0001 [1]

<span>There are Billions and billions of galaxies in the universe containing Trillions and trillions of stars in each galaxy.</span>

8 0

3 years ago

Read 2 more answers

A 500N person stands 2.5m from a wall against which a horizontal beam is attached. The beam is 6m long and weighs 200N. A cable

Vinil7 [7]

PART A)

Here by force balance along Y direction

$Tsin45 + F_y = 500 + 200$

Force balance along X direction

$Tcos45 = F_x$

now by torque balance

$Tsin45 (6) = 500 (2.5) + 200 (3)$

$T(4.24) = 1250 + 600$

$T = 436.3 N$

PART B)

now from above equations

$Tsin45 + F_y = 700$

$F_y = 391.5 N$

$F_x = Tcos45 = 308.5 N$

now net reaction force of wall is given as

$F = \sqrt{F_x^2 + F_y^2}$

$F = 498.4 N$

5 0

3 years ago

What is your displacement if you travel 5 miles north to the store

kumpel [21]

Answer:

5miles North

Explanation:

The displacement to the store is 5miles northward.

Displacement is the distance traveled in a specific direction. Displacement is a vector quantity.

A vector has both magnitude and direction.

The magnitude is the amount of that quantity

The direction is its orientation from a reference.

Therefore, the displacement is 5miles north

3 0

3 years ago

If The density of this stainless steel is7.85 g/cm3,specific heatis 0.5 J/g.K, melting pointis 1673K, heat of fusion s0.260J/kg.

kodGreya [7K]

Answer:

$\Delta H=687.4 J$

Explanation:

Hello!

In this case, for this melting process, we can identify two sub-processes in order to take the stainless steel from solid to liquid:

1. Heat up from 298.15 K to 1673 K.

2. Undergo the phase transition.

Both process have an associated enthalpy as shown below:

$\Delta H_1=1g*0.5\frac{J}{g*K} (1673K-298.15K)=687.4J$

$\Delta H_2=0.001kg*\frac{0.260J}{kg} =0.00026J$

Therefore, the required heat is:

$\Delta H=\Delta H_1+\Delta H_2\\\\\Delta H=687.4J+0.00026J\\\\\Delta H=687.4J$

Notice the problem is not providing neither the mass or volume, that is why we assumed the mass is 1 g; however, it can be changed to the mass you are given.

Best regards!

4 0

2 years ago

Can someone help me please

Svetllana [295]

Answer:

B.) Visit a healthcare professional

Explanation:

Their judgement is much better than a friend's or your own.

6 0

3 years ago