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3 years ago

10 Points

Physics

2 answers:

zaharov [31]3 years ago

8 0

D.radiative zone but I’m not really sure, I’m sorry if it’s wrong

Vinvika [58]3 years ago

4 0

D not positive though

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A 1200-kg ore cart is rolling at 10.8 m/s across a flat friction-free surface. a crane suddenly drops of ore vertically into the

Andre45 [30]

The value of the final speed depends on the mass of the ore.

Let's call m the mass of the ore. We can solve the exercise by requiring the conservation of momentum, which must be the same before and after the ore is loaded.

Initially, there is only the cart, so the momentum is
$p=Mv=(1200 kg)(10.8 m/s)=12960 kg m/s$
After the ore is loaded, the new mass will be (1200 kg+m), and the new speed is $v_f$ . The momentum p is conserved, so it is still 12960 kg m/s. Therefore, we have
$p=12960 kg m/s =(1200 kg+m)v_f$
and so the final speed is
$v_f = \frac{12960 kg m/s}{1200 kg +m}$

5 0

3 years ago

HOW DOES A GRAPH HELP YOU SEE ANOMALOUS DATA?

Ghella [55]

Answer:

Anomalies consist of one or more modifications, insertions or deletions. As was described in Section 3.1, there are only three types of changes that can be made to a graph. Therefore, anomalies that consist of structural changes to a graph must consist of one of these types. Assumption 4.

Explanation:

8 0

3 years ago

For a short time the missile moves along the parabolic path y=(18−2x2) km. If motion along the ground is measured as x=(4t−3) km

muminat

Answer with Explanation:

We are given that

$y=(18-2x^2) km$

$x=(4t-3)km$

Differentiate x and y w.r.t t

$\frac{dx}{dt}=4$

$\frac{dy}{dt}=-4x\frac{dx}{dt}$

$\frac{dy}{dt}=-4x\times 4=-16x=-16(4t-3)$

$v_x=\frac{dx}{dt}=4$

$v_y=\frac{dy}{dt}=-16(4t-3)$

Substitute t=1

$v_x=4$

$v_y=-16(4-3)=-16$

Magnitude of velocity= $\mid v\mid=\sqrt{v^2_x+v^2_y}$

$\mid v\mid=\sqrt{4^2+(-16)^2}=16.49 m/s$

Hence, the magnitude of the missile's velocity=16.49 m/s

$a_x=\frac{d(\frac{dx}{dt})}{dt}=\frac{d(4)}{dt}=0$

$a_y=\frac{d(\frac{dy}{dt})}{dt}=\frac{d(-16(4t-3))}{dt}=-64$

Substitute t=1

$a_x=0,a_y=-64$

$\mid a\mid=\sqrt{a^2_x+a^2_y}$

$\mid a\mid=\sqrt{0+(-64)^2}=64m/s^2$

Hence, the magnitude of acceleration when t=1 s= $64m/s^2$

3 0

2 years ago

A pendulum consisting of a 0.5 kg mass tied to a 0.3 m string is set into oscillation at the same moment that a stone is dropped

lara31 [8.8K]

Answer:

2.72 cycles

Explanation:

First of all, let's find the time that the stone takes to reaches the ground. The stone moves by uniform accelerated motion with constant acceleration g=9.8 m/s^2, and it covers a distance of S=44.1 m, so the time taken is

$S=\frac{1}{2}at^2\\t=\sqrt{\frac{2S}{a}}=\sqrt{\frac{2(44.1m)}{9.8 m/s^2}}=3 s$

The period of the pendulum instead is given by:

$T=2 \pi \sqrt{\frac{L}{g}}=2 \pi \sqrt{\frac{0.3 m}{9.8 m/s^2}}=1.10 s$

Therefore, the number of oscillations that the pendulum goes through before the stone hits the ground is given by the time the stone takes to hit the ground divided by the period of the pendulum:

$N=\frac{t}{T}=\frac{3 s}{1.10 s}=2.72$

6 0

3 years ago

A weight suspended from a spring is seen to bob up and down over a distance of 20 cm triply each second. What is the period? Wha

Kisachek [45]

Answer:

1) Hence, the period is 0.33 s.

2) The amplitude is 10 cm.

Explanation:

1) The period is given by:

$T = \frac{1}{f}$

Where:

f: is the frequency = 3 bob up and down each second = 3 s⁻¹ = 3 Hz

$T = \frac{1}{f} = \frac{1}{3 Hz} = 0.33 s$

Hence, the period is 0.33 s.

2) The amplitude is the distance between the equilibrium position and the maximum position traveled by the spring. Since the spring is moving up and down over a distance of 20 cm, then the amplitude is:

$A = \frac{20 cm}{2} = 10 cm$

Therefore, the amplitude is 10 cm.

I hope it helps you!

5 0

3 years ago