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3 years ago

Answer this question please and you will can’t the brain list

Physics

1 answer:

Eduardwww [97]3 years ago

4 0

Answer:

We know that velocity equals to difference in distance divided by difference in time.

1st period: velocity is increasing. Where v = (8-0)/(3-0) = 8/3 m/s.

2nd one: velocity decreases. Where

v=(5-8)/(4-3) = (-3) m/s.

3rd one: velocity is constant because no change in distance, so v=0.

4th one: velocity increases again. Where v=(10-5)/(8-6)= 5/2 m/s.

5th one: velocity is constant again, so v=0.

6th one: velocity decreases, where

v=(6-10)/(15-12)= (-4)/3 m/s.

You might be interested in

Light of wavelength 436.1 nm falls on two slits spaced 0.31 mm apart. What is the required distance from the slits to the screen

kakasveta [241]

Answer:

The correct answer is "4.26 m".

Explanation:

Given:

Wavelength,

$\lambda = 436.1 \ nm$

or,

$=436.1\times 10^{-9} \ m$

Distance,

$d = 0.31 \ mm$

or,

$=0.31\times 10^{-3} \ m$

Distance between the 1st and 2nd dark fringes,

$(y_2-y_1) = 6\times 10^{-3} \ m$

As we know,

⇒ $\frac{d}{L} (y_2-y_1) = \lambda$

or,

⇒ $L=\frac{d(y_2-y_1)}{\lambda}$

By substituting the values, we get

$=\frac{0.31\times 6\times 10^{-6}}{436.1\times 10^{-9}}$

$=\frac{0.31\times 6\times 10^3}{436.1}$

$=\frac{1860}{436.1}$

$=4.26 \ m$

3 0

3 years ago

1. It s defficult to lift a heavier stone than the lighter one. why

stiks02 [169]

Answer:

lThe effect of the attraction of the earth on a bigger stone can be observed more than the effect of attraction of the earth on a smaller one. hence it is difficult to lift a large stone than the smaller one on the earth surface.

4 0

3 years ago

A pendulum with a mass of 250 g is released from its maximum displacement of 5 degrees and swings with a simple harmonic motion.

Amiraneli [1.4K]

L=T2xg/4x3.14^2
16x9.8/4x3.14^2
Used your calculator plz

3 0

3 years ago

Three blocks are placed in contact on a horizontal frictionless surface. A constant force of magnitude F is applied to the box o

Lina20 [59]

Answer:

A) M

Explanation:

The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:

Box with mass M

$\Sigma F = F - F' = M\cdot a$

Box with mass 2M

$\Sigma F = F' - F'' = 2\cdot M \cdot a$

Box with mass 3M

$\Sigma F = F'' = 3\cdot M \cdot a$

On the third equation, acceleration can be modelled in terms of F'':

$a = \frac{F''}{3\cdot M}$

An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.

$F' = 2\cdot M \cdot a + F''$

$F' = 2\cdot M \cdot \left(\frac{F''}{3\cdot M} \right) + F''$

$F' = \frac{5}{3}\cdot F''$

Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:

$F - \frac{5}{3}\cdot F'' = M \cdot \left(\frac{F''}{3\cdot M} \right)$

$F = \frac{5}{3} \cdot F'' + \frac{1}{3}\cdot F''$

$F = 2\cdot F''$

$F'' = \frac{1}{2}\cdot F$

Afterwards, F' as function of the external force can be obtained by direct substitution:

$F' = \frac{5}{6}\cdot F$

The net forces of each block are now calculated:

Box with mass M

$M\cdot a = F - \frac{5}{6}\cdot F$

$M\cdot a = \frac{1}{6}\cdot F$

Box with mass 2M

$2\cdot M\cdot a = \frac{5}{6}\cdot F - \frac{1}{2}\cdot F$

$2\cdot M \cdot a = \frac{1}{3}\cdot F$

Box with mass 3M

$3\cdot M \cdot a = \frac{1}{2}\cdot F$

As a conclusion, the box with mass M experiments the smallest net force acting on it, which corresponds with answer A.

8 0

4 years ago

A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab (Fig. 6-58). The coefficient of static fricti

vovikov84 [41]

Answer:

<u> $\text { The "resulting action" on the slab is } 0.98 \mathrm{m} / \mathrm{s}^{2}$ </u>

Explanation:

Normal reaction from 40 kg slab on 10 kg block

M × g = 10 × 9.8 = 98 N

Static frictional force = 98 × 0.7 N

Static frictional force = 68.6 N is less than 100 N applied

10 kg block will slide on 40 kg slab and net force on it

= 100 N - kinetic friction

$=100-(98 \times 0.4)\left(\mu_{\text {kinetic }}=0.4\right)$

= 100 - 39.2

= 60.8 N

$10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with, } \frac{\mathrm{Net} \text { force }}{\text { mass }}$

$10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with }=\frac{60.8}{10}$

$10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with }=6.08 \mathrm{m} / \mathrm{s}^{2}$

$\text { Frictional force on 40 kg slab by 10 kg block, normal reaction \times \mu_{kinetic } }$

Frictional force on 40 kg slab by 10 kg block = 98 × 0.4

Frictional force on 40 kg slab by 10 kg block = 39.2 N

$40 \mathrm{kg} \text { slab will move with } \frac{\text { frictional force }}{\text { mass }}$

$40 \mathrm{kg} \text { slab will move with }=\frac{39.2}{40}$

40 kg slab will move with = $0.98 \mathrm{m} / \mathrm{s}^{2}$

$\text { The "resulting action" on the slab is } 0.98 \mathrm{m} / \mathrm{s}^{2}$

3 0

3 years ago