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3 years ago

7

A volcano launches a lava bomb straight upward with an initial speed of 24 m/s. speed at 2 and 3 seconds and it it is upward or

downward

1 answer:

Serggg [28]3 years ago

8 0

Answer:

Explanation:

v = u + at

Let Up be the positive direction

v(2) = 24 + (-9.8)(2) = 4.4 m/s Positive result means Upward

v(3) = 24 + (-9.8)(3) = -5.4 m/s Negative result means Downward

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A 50 gram meterstick is placed on a fulcrum at its 50 cm mark. A 20 gram mass is attached at the 12 cm mark. Where should a 40 g

alekssr [168]

Answer:

The 40g mass will be attached at 69 cm

Explanation:

First, make a sketch of the meterstick with the masses placed on it;

--------------------------------------------------------------------------

↓ Δ ↓

20 g.................50 cm.................40g

38 cm y cm

Apply principle of moment;

sum of clockwise moment = sum of anticlockwise moment

40y = 20 (38)

40y = 760

y = 760 / 40

y = 19 cm

Therefore, the 40g mass will be attached at 50cm + 19cm = 69 cm

12cm 50 cm 69cm

--------------------------------------------------------------------------

↓ Δ ↓

20 g.................50 cm.................40g

38 cm 19 cm

5 0

3 years ago

Why does changing the shape of an object have no effect on the density?

iren [92.7K]

Because the object is still made of the same material
Density is not affected by the weight and shape of an object its affected by how concentrated the atoms are in a given volume

7 0

3 years ago

You slide a chair across a rough, horizontal surface. The chair's mass is 18.8 kg. The force you exert on the chair is 156 N dir

Luda [366]

Answer:

626.612 J

Explanation:

Work done by friction on the chair is given as

W-ΔEk = W'..................... Equation 1

Where W' = Work done by friction on the chair, W = Work done on the chair by me, Ek = change in Kinetic energy of the chair as a result of the slide.

From the question,

W = FdcosФ.............. Equation 2

ΔEk = 1/2m(v²-u²)................ Equation 3

Where F = Force applied on the chair, d = distance of slide, Ф = angle between the force and the horizontal, m = mass of the chair, v = final velocity of the chair, u = initial velocity of the chair

Substitute equation 2 and equation 3 into equation 1

W' = FdcosФ-1/2m(v²-u²)........................ Equation 4

Given: F = 156, d = 5 m, Ф = 26°, m = 18.8 kg, v = 3.1 m/s, u = 1.3 m/s

Substitute into equation 4

W' = 156×5×cos26°-1/2×18(3.1²-1.3²)

W' = 701.06-74.448

W' = 626.612 J.

Hence the work done by friction = 626.612 J

8 0

4 years ago

A wave with a frequency of 1200 Hz propagates along a wire that is under a tension of 800 N. Its wavelength is 39.1 cm. What wil

mash [69]

Answer:

The wavelength will be 33.9 cm

Explanation:

Given;

frequency of the wave, F = 1200 Hz

Tension on the wire, T = 800 N

wavelength, λ = 39.1 cm

$F = \frac{ \sqrt{\frac{T}{\mu} }}{\lambda}$

Where;

F is the frequency of the wave

T is tension on the string

μ is mass per unit length of the string

λ is wavelength

$\sqrt{\frac{T}{\mu} } = F \lambda\\\\\frac{T}{\mu} = F^2\lambda^2\\\\\mu = \frac{T}{F^2\lambda^2} \\\\\frac{T_1}{F^2\lambda _1^2} = \frac{T_2}{F^2\lambda _2^2} \\\\\frac{T_1}{\lambda _1^2} = \frac{T_2}{\lambda _2^2}\\\\T_1 \lambda _2^2 = T_2\lambda _1^2\\\\$

when the tension is decreased to 600 N, that is T₂ = 600 N

$T_1 \lambda _2^2 = T_2\lambda _1^2\\\\\lambda _2^2 = \frac{T_2\lambda _1^2}{T_1} \\\\\lambda _2 = \sqrt{\frac{T_2\lambda _1^2}{T_1}} \\\\\lambda _2 = \sqrt{\frac{600* 0.391^2}{800}}\\\\\lambda _2 = \sqrt{0.11466} \\\\\lambda _2 =0.339 \ m\\\\\lambda _2 =33.9 \ cm$

Therefore, the wavelength will be 33.9 cm

5 0

3 years ago

Youare on a train travelling northat 80.0m/s relative to the ground. The air is still relativetothegroundwhen you hear the whist

AURORKA [14]

To solve this problem we will apply the concepts related to the Doppler effect. This is understood as the change in apparent frequency of a wave produced by the relative movement of the source with respect to its observer. Mathematically this is given as,

$f = \frac{v \pm v_r}{v \pm v_s}(f_0)$

Here,

v = Speed of the waves in the middle

$v_r$ = Speed of the receiver in relation to the medium (Positive if the receiver is moving towards the transmitter or vice versa)

$v_s$ = Speed of the source with respect to the medium (Positive if the source moves away from the receiver or vice versa)

Our values are given as,

$v = 342m/s$

$f_0 = 262Hz$

$v_r = 80m/s$

$f = 350Hz$

Replacing,

$350 = \frac{342+80}{342-v} (262)$

Solving for the velocity of the source,

$v = 26.1m/s$

Therefore the speed of the other train is 26.1m/s

3 0

3 years ago