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3 years ago

Use the graph above to determine the change in speed of the object between 20 and 30 seconds?

Physics

2 answers:

allsm [11]3 years ago

4 0

Answer:

6 m/s

Explanation:

To determine the change in speed of the object, we just need to determine its speed at t = 30 s and at t = 20 s, and then calculate the difference.

The speed at t = 30 is:

v = 6 m/s

While the speed at t = 20 s is:

u = 0

Therefore, the change in speed is:

$\Delta v = v-u=6-0 = 6 m/s$

lara31 [8.8K]3 years ago

4 0

Explanation:

The attached figure shows the acceleration time graph of an object. We known that the slope of the a-t graph gives the velocity of an object. The area under the a-t graph gives the velocity of object.

In x axis, time t = 30 s- 20 s = 10 s

In y axis, acceleration, $a = 6 - 0=6\ m/s^2$

Area under the graph from 20 to 30 seconds is given by :

$A=v=\dfrac{1}{2}\times 10\times 6$

v = 30 m/s

So, the change in speed of the object is 30 m/s. Hence, this is the required solution.

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A new prototype cup has been designed to keep liquids, such as hot coffee or cold annk, near their original temperature for long

san4es73 [151]

Answer:

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3 years ago

At t=0, a block A of mass 8 kg and block B of mass 16 kg are both at position x=0 . Block A is at rest, and block B is moving at

love history [14]

The center of mass of the two objects is the average position of the parts of the two object system

The center of mass of block A, and block B after displacement of block B is at 20 m from block A

Reason:

The given parameters are;

The position of block A and block B at t = 0 is x = 0

The mass of block A, m₁ = 8 kg

Mass of block B, m₂ = 16 kg

Speed of block A = 0 m/s

Speed of block B, v₂ = 10 m/s

Location of the center of mass of the two object at t = 3 s; Required

Solution;

The location of block A, after 3 s is x₁ = 0 (block A is at rest)

The location of block B, = v₂ × t

The location of block B, after 3 s is x₂ = 10 m/s × 3 s = 30 m

The center of mass of two masses are given as follows;

$x_{cm} = \dfrac{m_1 \cdot x_1 +m_2\cdot x_2}{m_1 + m_2}$

$x_{cm} = \dfrac{8 \times0 + 16 \times 30}{8 + 16} = 20$

The center of mass of the two objects is at at the position x = 20 m (from block A)

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3 years ago

An eccentric inventor attempts to levitate a cork ball by wrapping it with foil and placing a large negative charge on the ball

djverab [1.8K]

Answer:

because of the idea that like charges get repulsion as a force.

Explanation:

because you wrap the ball with foil, the negative charges will leave the foil and go into the ball by induction. This leaves the foil as a positively charged particle since its electrons left it for the ball, making the ball a negatively charged particle. but if you bring the negative charge near the foil, the electrons will transfer from that and go into the foil, making it negatively charged. Now, because the ball and the foil have the same charge, they repel. the foil flies off.

7 0

3 years ago

Consider a mechanical clutch that consists of two heavy disks that can engage or disengage. At the beginning disk 1 with mass m1

miv72 [106K]

Answer:

w = 4,786 rad / s , f = 0.76176 Hz

Explanation:

For this problem let's use the concept of angular momentum

L = I w

The system is formed by the two discs, during the impact the system remains isolated, we have the forces are internal, this implies that the external torque is zero and the angular momentum is conserved

Initial Before sticking

L₀ = 0 + I₂ w₂

Final after coupling

$L_{f}$ = (I₁ + I₂) w

The moments of inertia of a disk with an axis of rotation in its center are

I = ½ M R²

How the moment is preserved

L₀ = $L_{f}$

I₂ w₂ = (I₁ + I₂) w

w = w₂ I₂ / (I₁ + I₂)

Let's reduce the units to the SI System

d₁ = 60 cm = 0.60 m

d₂ = 40 cm = 0.40 m

f₂ = 200 min-1 (1 min / 60 s) = 3.33 Hz

Angular velocity and frequency are related.

w₂ = 2 π f₂

w₂ = 2π 3.33

w₂ = 20.94 rad / s

Let's replace

w = w₂ (½ M₂ R₂²) / (½ M₁ R₁² + ½ M₂ R₂²)

w = w₂ M₂ R₂² / (M₁ R₁² + M₂ R₂²)

Let's calculate

w = 20.94 8 0.40² / (12 0.60² + 8 0.40²)

w = 20.94 1.28 / 5.6

w = 4,786 rad / s

Angular velocity and frequency are related.

w = 2π f

f = w / 2π

f = 4.786 / 2π

f = 0.76176 Hz

7 0

3 years ago

An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.7

Kazeer [188]

Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

$v = \frac{I}{nqA}$

Where:

I: is the current = 4.50 A

n: is the number of electrons

q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C

A: is the cross-sectional area = 2.20x10⁻⁶ m²

We need to find the number of electrons:

$n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3}$

Now, we can find the drift speed:

$v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s$

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!

4 0

3 years ago