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2 years ago

5

An object travels 50 m in 4 s. It had no initial velocity and experiences constant acceleration. What is the magnitude of the ac

celeration?

1 answer:

Allisa [31]2 years ago

3 0

Answer:

The formula to calculate velocity in this case:

v = v0 + at

=> a = (v - v0)/t

= (50 - 0)/4

= 50/4 = 12.5 (m/s2)

Hope this helps!

:)

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A block of mass 3m is placed on a frictionless horizontal surface, and a second block of mass m is placed on top of the first bl

tatuchka [14]

By Newton's second law, assuming <em>F</em> is horizontal,

• the net <u>horizontal</u> force on the <u>larger</u> block is

<em>F</em> - <em>µmg</em> = 3<em>mA</em>

where <em>µmg</em> is the magnitude of friction felt by the larger block due to rubbing with the smaller one, <em>µ</em> is the coefficient of static friction between the two blocks, and <em>A</em> is the block's acceleration;

• the net <u>vertical</u> force on the <u>larger</u> block is

4<em>mg</em> - 3<em>mg</em> - <em>mg</em> = 0

where 4<em>mg</em> is the mag. of the normal force of the surface pushing up on the combined mass of the two blocks, 3<em>mg</em> is the weight of the larger block, and <em>mg</em> is the weight of the smaller block;

• the net <u>horizontal</u> force on the <u>smaller</u> block is

<em>µmg</em> = <em>ma</em>

where <em>µmg</em> is again the friction between the two blocks, but notice that this points in the same direction as <em>F</em>. It is the only force acting on the smaller block in the horizontal direction, so (b) static friction is causing the smaller block to accelerate;

• the net <u>vertical</u> force on the <u>smaller</u> block is

<em>mg</em> - <em>mg</em> = 0

where <em>mg</em> is the magnitude of both the normal force of the larger block pushing up on the smaller one, and the weight of the smaller block.

(You should be able to draw your own FBD's based on the forces mentioned above.)

(c) Solve the equations above for <em>A</em> and <em>a</em> :

<em>A</em> = (<em>F</em> - <em>µmg</em>) / (3<em>m</em>)

<em>a</em> = <em>µg</em>

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1 year ago

The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine th

Luba_88 [7]

Here is the complete question

The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine the horizontal shear stress at point H, which is located L = 20 mm below the centriod

The missing image which is the remaining part of this question is attached in the image below.

Answer:

The horizontal shear stress at point H is $\mathbf{\tau_H \approx 42.604 \ N/mm^2}$

Explanation:

Given that :

The internal shear force V = 80 kN = 80 × 10³ N

The moment of inertia = 64,900,000

The length = 20 mm below the centriod

The horizontal shear stress $\tau$ can be calculated by using the equation:

$\tau = \dfrac{VQ}{Ib}$

where;

Q = moment of area above or below the point H

b = thickness of the beam = 10 mm

From the centroid ;

Q = $Q_1 + Q_{2}$

Q = $A_1y_1 + A_{2}y_{2}$

Q = ( ( 70 × 10) × (55) + ( 210 × 15) (90 + 15/2) ) mm³

Q = ( ( 700) × (55) + ( 3150 ) ( 97.5) ) mm³

Q = ( 38500 + 307125 ) mm³

Q = 345625 mm³

$\tau_H = \dfrac{VQ}{Ib}$

$\tau_H = \dfrac{80*10^3 * 345625}{64900000*10 }$

$\tau_H = \dfrac{2.765*10^{10}}{649000000 }$

$\tau_H = 42.60400616 \ N/mm^2$

$\mathbf{\tau_H \approx 42.604 \ N/mm^2}$

The horizontal shear stress at point H is $\mathbf{\tau_H \approx 42.604 \ N/mm^2}$

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2 years ago

A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur

Nonamiya [84]

Answer:

Check the explanation

Explanation:

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

The image is virtual

The image is upright

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

Kindly check the diagram in the attached image below.

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1 year ago

A solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ωi. The sphere is s

Leona [35]

Answer:

0.6

Explanation:

The volume of a sphere = $\frac{4}{3} \pi (\frac{D}{2})^3$

Therefore $\pi * r^2 * (\frac{D}{2} ) = \frac{4}{3} \pi (\frac{D}{2})^3$

r of the disc = $1.15(\frac{ D}{2} )$

Using conservation of angular momentum;

The $M_i$ of the sphere = $\frac{2}{5} m \frac{D}{2}^2$

$M_i$ of the disc = $m*\frac{ \frac{1.15*D}{2}^2 }{2}$

$\frac{wd}{ws} = \frac{\frac{2}{5}m * \frac{D}{2}^2}{ m * \frac{(\frac{`.`5*D}{2})^2 }{2} }$

= 0.6

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2 years ago

Water is to be boiled at sea level (1 atm pressure) in a 30-cm-diameter stainless steel pan placed on top of a 18-kW electric bu

Tamiku [17]

Answer:

Explanation:

18 kW = 18000 J /s

60% of 18kW = 10800 J/s

Latent heat of evaporation of water

= 2260 x 10³ J / kg

kg of water being evaporated per second

= 10800 / 2260 x 10³ kg /s

= 4.7787 x 10⁻³ kg / s

= 4.78 gm / s .

3 0

1 year ago