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3 years ago

A small object is attached to a horizontal spring and set in simple harmonic motion with amplitude A and period T .

Physics

1 answer:

kupik [55]3 years ago

4 0

Answer:

t = 3/2T

To find how long it takes to cover a total distance of 6A, we need to find the time it takes to cover a distance A then multiply by 6.

The step to the solution is given below in the attachment.

Explanation:

Thank you for reading

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A running back with a mass of 70 kg travels down the field with a velocity of 5.0 m s . Calculate the kinetic energy of the foot

jarptica [38.1K]

K.E = 1/2 mv²

= 1/2 (70kg) (5.0ms)²

= 1 ˣ ( 1750) / 2

= 875 kgms²

3 0

3 years ago

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Indicate on the chart whether you would classify each model as representing an element, compound, or mixture.

Ne4ueva [31]

Answer:

1 compound

2 mixture

3 elements

4 0

3 years ago

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A large asteroid of mass 98700 kg is at rest far away from any planets or stars. A much smaller asteroid, of mass 780 kg, is in

Citrus2011 [14]

Answer:

1.81 x 10^-4 m/s

Explanation:

M = 98700 kg

m = 780 kg

d = 201 m

Let the speed of second asteroid is v.

The gravitational force between the two asteroids is balanced by the centripetal force on the second asteroid.

$\frac{GMm}{d^{2}}=\frac{mv^2}{d}$

$v=\sqrt{\frac{GM}{d}}$

Where, G be the universal gravitational constant.

G = 6.67 x 10^-11 Nm^2/kg^2

$v=\sqrt{\frac{6.67 \times 10^{-11}\times 98700}{201}}$

v = 1.81 x 10^-4 m/s

7 0

3 years ago

An archer fires an arrow at a castle 230 m away. The arrow is in flight for 6 seconds, then hits the wall of the castle and stic

Vika [28.1K]

Answer:

(a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

Explanation:

Given that,

Horizontal distance = 230 m

Time t = 6 sec

Vertical distance = 16 m

We need to calculate the horizontal component

Using formula of horizontal component

$R =u\cos\theta t$

Put the value into the formula

$\dfrac{230}{6} = u\cos\theta$

$u\cos\theta=38.33$ .....(I)

We need to calculate the height

Using vertical component

$H=u\sin\theta t-\dfrac{1}{2}gt^2$

Put the value in the equation

$16 =u\sin\theta\times6-\dfrac{1}{2}\times9.8\times6^2$

$u\sin\theta=\dfrac{16+9.8\times18}{6}$

$u\sin\theta=32.06$ .....(II)

Dividing equation (II) and (I)

$\dfrac{u\sin\theta}{u\cos\theta}=\dfrac{32.06}{38.33}$

$\tan\theta=0.8364$

$\theta=\tan^{-1}0.8364$

$\theta=39.90^{\circ}$

(a). We need to calculate the initial speed

Using equation (I)

$u\cos\theta\times t=38.33$

Put the value into the formula

$u =\dfrac{230}{6\times\cos39.90}$

$u=49.96\ m/s$

(b). We have already calculate the angle.

Hence, (a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

5 0

3 years ago

A skydiver of 75 kg mass has a terminal velocity of 60 m/s. At what speed is the resistive force on the skydiver half that when

ankoles [38]

Answer:

The speed of the resistive force is 42.426 m/s

Explanation:

Given;

mass of skydiver, m = 75 kg

terminal velocity, $V_T = 60 \ m/s$

The resistive force on the skydiver is known as drag force.

Drag force is directly proportional to square of terminal velocity.

$F_D = kV_T^2$

Where;

k is a constant

$k = \frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2}$

When the new drag force is half of the original drag force;

$F_D_2 = \frac{F_D_1}{2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_1}{2V_{T2}^2} \\\\\frac{1}{V_{T1}^2} = \frac{1}{2V_{T2}^2}\\\\2V_{T2}^2 = V_{T1}^2\\\\V_{T2}^2= \frac{V_{T1}^2}{2} \\\\V_{T2}= \sqrt{\frac{V_{T1}^2}{2} } \\\\V_{T2}= \frac{V_{T1}}{\sqrt{2} } \\\\V_{T2}= 0.7071(V_{T1})\\\\V_{T2}= 0.7071(60 \ m/s)\\\\V_{T2}= 42.426 \ m/s$

Therefore, the speed of the resistive force is 42.426 m/s

8 0

3 years ago