A permanent magnet

1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp

Tanzania [10]

Answer with explanation:

We are given that

Mass of ball, $m_1=$ 75 g= $\frac{75}{1000}=0075kg$

1 kg=1000 g

Height, $h_1=1.6 m$

$h_2=0.6 m$

Horizontal velocity, $v_x=2 m/s$

Mass of plate $m_2=400 g=\frac{400}{1000}=0.4 kg$

a.Initial velocity of plate, $u_2=0$

Velocity before impact= $u_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 1.6}=5.6m/s$

Where $g=9.8 m/s^2$

Velocity after impact, $v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s$

According to law of conservation of momentum

$m_1u_1+m_2u_1=-m_1v_1+m_2v_2$

Substitute the values

$0.075\times 5.6+0=-0.075\times 3.4+0.4v_2$

$0.4v_2=0.075\times 5.6+0.075\times 3.4$

$v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s$

Velocity of plate=1.69 m/s

b.Initial energy= $\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J$

Final energy= $\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2$

Final energy= $\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J$

Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

6 0

3 years ago

A 500 kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 30 N/m. The blo

m_a_m_a [10]

Answer:

x = 0.396 m

Explanation:

The best way to solve this problem is to divide it into two parts: one for the clash of the putty with the block and another when the system (putty + block) compresses it is spring

Data the putty has a mass m1 and velocity vo1, the block has a mass m2 . t's start using the moment to find the system speed.

Let's form a system consisting of putty and block; For this system the forces during the crash are internal and the moment is preserved. Let's write the moment before the crash

p₀ = m1 v₀₁

Moment after shock

$p_{f}$ = (m1 + m2) $v_{f}$

p₀ = $p_{f}$

m1 v₀₁ = (m1 + m2) $v_{f}$

$v_{f}$ = v₀₁ m1 / (m1 + m2)

$v_{f}$ = 4.4 600 / (600 + 500)

$v_{f}$ = 2.4 m / s

With this speed the putty + block system compresses the spring, let's use energy conservation for this second part, write the mechanical energy before and after compressing the spring

Before compressing the spring

Em₀ = K = ½ (m1 + m2) $v_{f}$ ²

After compressing the spring

$E_{mf}$ = Ke = ½ k x²

As there is no rubbing the energy is conserved

Em₀ = $E_{mf}$

½ (m1 + m2) $v_{f}$ ² = = ½ k x²

x = $v_{f}$ √ (k / (m1 + m2))

x = 2.4 √ (11/3000)

x = 0.396 m

7 0

4 years ago

Suppose an object is moving in a straight line at 50 miles/hr. According to Newton's first law of motion, the object will ______

Setler [38]

Answer:

B

Explanation:

Newton's first law of motion states that a body will remain in its state of rest or if its in motion will continue to move in a straight line, unless its acted upon by an external force.The ability of an object to stay at rest or in motion if its in motion is known as inertia.

Hence the correct option is B.

8 0

4 years ago

A car of mass 2400kg moving at 20 m/s slams into a cement wall and comes to a halt?

Brut [27]

Impulse = change in momentum

The car's momentum was (mass) x (speed)

Momentum = (2400 kg) x (20 m/s)

Momentum = 48,000 km-m/s

To completely stop the car, the impulse = -48,000 km-m/s .

7 0

4 years ago

A mineral has a mas of 53grams. Its volume is 12millileters. What is the<br> density?

polet [3.4K]

Answer:

636

Explanation:

6 0

4 years ago

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