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3 years ago

Find your average speed if you run 50m in 10s

Physics

2 answers:

attashe74 [19]3 years ago

8 0

Average speed = (distance covered) / (time to cover the distance)

Average speed = (50 m) / (10 s)

Average speed = 50/10 m/s

Average speed = 5 m/s

Anastasy [175]3 years ago

6 0

50m divided by 10s equals 5m/s.
You're welcome.

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The 10-kg block is held at rest on the smooth inclined plane by the stop block at A. If the10-g bullet is traveling at 300m/swhe

WARRIOR [948]

Answer:

6.8 mm

Explanation:

We are given that

Mass of block,m=10 kg

Mass of bullet, $m_b=10 g=10\times 10^{-3} kg$

1 kg=1000 g

Total mass of system,M= $m+m_a=10+10\times 10^{-3}=10.01kg$

Speed of bullet,u=300 m/s

$\theta=30^{\circ}$

By law of conservation of momentum

$m_bucos\theta=Mv$

$v=\frac{m_bvcos\theta}{M}=\frac{0.01\times 300cos30^{\circ}}{10.01}=0.259m/s$

According to law of conservation of energy

Change in kinetic energy of system=Change in potential energy of system

$\frac{1}{2}Mv^2-0=Mgh-0$

$\frac{1}{2}(10.01)(0.259)^2=10.01\times 9.8 h$

Where $g=9.8 m/s^2$

$h=\frac{(0.259)^2}{2\times 9.8}=0.0034m$

1m=100 cm

$h=0.0034\times 100=0.34 cm$

Distance traveled by block= $d=\frac{h}{sin\theta}=\frac{0.34}{sin30^{\circ}}=0.68 cm=6.8 mm$

1cm=10 mm

4 0

4 years ago

A 0.540-kg bucket rests on a scale. Into this bucket you pour sand at the constant rate of 56.0 g/s. If the sand lands in the bu

romanna [79]

Answer:

a) 12.8212 N

b) 12.642 N

Explanation:

Mass of bucket = m = 0.54 kg

Rate of filling with sand = 56.0 g/ sec = 0.056 kg/s

Speed of sand = 3.2 m/s

g= 9.8 m/sec2

<u>Condition (a);</u>

Mass of sand = Ms = 0.75 kg

So total mass becomes = bucket mass + sand mass = 0.54 +0.75=1.29 kg

== > total weight = 1.29 × 9.8 = 12.642 N

Now impact of sand = rate of filling × velocity = 0.056 × 3.2 = 0.1792 kg. m /sec2=0.1792 N

Scale reading is sum of impact of sand and weight force ;

i-e

scale reading = 12.642 N+0.1792 N = 12.8212 N

<u>Codition (b);</u>

bucket mass + sand mass = 0.54 +0.75=1.29 kg

==>weight = mg = 1.29 × 9.8 = 12.642 N (readily calculated above as well)

6 0

3 years ago

For the last part of the lab, you should have found the mass of the meter stick. So if a mass of 85 g was placed at the 2 cm MAR

Bond [772]

Answer:

272.89g

Explanation:

Find the diagram to the question in the attachment below;.

Using the principle of moment to solve the question which states that the sum of clockwise moment is equal to the sum of anticlockwise moment.

Moment = Force * Perpendicular distance

Taking the moment of force about the pivot.

Anticlockwise moment:

The 85g mass will move in the anticlockwise

Moment of 85g mass = 85×36.6

= 3111gcm

Clockwise moment.

The mass of the metre stick M situated at the centre (50cm from each end) will move in the clockwise direction towards the pivot.

CW moment = 11.4×M = 11.4M

Equating CW moment to the ACW moment we will have;

11.4M = 3111

M = 3111/11.4

M = 272.89g

The mass of the metre stick is 272.89g

5 0

4 years ago

A puddle of water has a thin film of gasoline floating on it. A beam of light is shining perpendicular on the film. If the wavel

g100num [7]

Answer:

option A

Explanation:

Given,

wavelength of light, $\lambda = 560\ nm$

refractive index of gasoline, n₁ = 1.40

Refractive index of water, n₂ = 1.33

thickness of the film, t = ?

Condition of constructive interference is given by

$2 n t = (m+\dfrac{1}{2})\lambda$

For minimum thickness of the film m = 0

From the question we can clearly observe that phase change from gasoline to air

so, n = 1.4

$2 \times 1.4 \times t = \dfrac{560}{2}$

$t = 100\ nm$

Hence, the correct answer is option A

7 0

4 years ago

What name is given to the wall of water that makes landfall just ahead of a hurricane?

andreev551 [17]

Pretty sure its a. storm surge

8 0

4 years ago