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2 years ago

You pushes a 25kg box with 40 N of force, but the friction on the crate is 15 N. What is the net

Physics

1 answer:

grandymaker [24]2 years ago

4 0

Answer:

25 N in the direction of your push

Explanation:

The friction force acts against your pushing force....it reduces the net amount of force you are exerting

Net force = 40 - 15 = 25 N

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Answer:

Explanation:

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3 years ago

A sanding disk with rotational inertia 3.8 x 10-3 kg·m2 is attached to an electric drill whose motor delivers a torque of magnit

Elis [28]

Answer:

(a) Angular momentum of disk is $1.343\, \frac{kg.m^{2}}{s}$

(b) Angular velocity of the disk is $353\frac{rad}{s}$

Explanation:

Given

Rotational inertia of the disk , $I=3.8\times 10^{-3}kg.m^{2}$

Torque delivered by the motor , $\tau =17N.m$

Torque is applied for duration , $\Delta t=79ms=0.079s$

(a)

Magnitude of angular momentum of the disk = Angular impulse produced by the torque

$\therefore L=\tau \Delta t=17\times 0.079\frac{kg.m^{2}}{s}$

=> $L=1.343\, \frac{kg.m^{2}}{s}$

Thus angular momentum of disk is $1.343\, \frac{kg.m^{2}}{s}$

(b)

Since Angular momentum , $L=I\omega$

where $\omega$ = Angular velocity of the disk

$=>\omega =\frac{L}{I}=\frac{1.343}{3.8\times 10^{-3}}\frac{rad}{s}$

$\therefore \omega =353\frac{rad}{s}$

Thus angular velocity of the disk is $353\frac{rad}{s}$

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3 years ago

A woman fires a rifle with barrel length of 0.5400 m. Let (0, 0) be where the 125 g bullet begins to move, and the bullet travel

dybincka [34]

Answer:

Explanation:

Given that,

Length of barrel =0.54m

Mass of bullet=125g=0.125kg

Force extend

F=16,000+10,000x-26,000x²

a. Work done is given as

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=0.54m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=0.54m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=0.54m

W= 16,000(0.54) + 5000(0.54²) - 8666.667(0.54³) +0+0-0

W=8640+1458-1364.69

W=8733.31J

The workdone by the gas on the bullet is 8733.31J

b. Work done is given as

Work done when the length=1.05m

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=1.05m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=1.05m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=1.05mm

W= 16,000(1.05) + 5000(1.05²) - 8666.667(1.05³) +0+0-0

W=16800+5512.5-10032.75

W=12,279.75J

The workdone by the gas on the bullet is 12,279.75J

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3 years ago

A 12-g wad of sticky clay is hurled horizontally at a 100-g wooden block initially at rest on a horizontal surface. The clay sti

klasskru [66]

Answer:

91.5 m/s

Explanation:

m = mass of clay = 12 g = 0.012 kg

M = mass of wooden block = 100 g = 0.1 kg

d = distance traveled by the combination before coming to rest = 7.5 m

μ = Coefficient of friction = 0.65

V = speed of the combination of clay and lock just after collision

V' = final speed of the combination after coming to rest = 0 m/s

acceleration caused due to friction is given as

a = - μ g

a = - (0.65) (9.8)

a = - 6.37 m/s²

Using the kinematics equation

V'² = V² + 2 a d

0² = V² + 2(- 6.37) (7.5)

V = 9.8 m/s²

v = speed of clay just before collision

Using conservation of momentum

m v = (m + M) V

(0.012) v = (0.012 + 0.100) (9.8)

v = 91.5 m/s

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3 years ago

A plastic rod of length d = 1.5 m lies along the x-axis with its midpoint at the origin. The rod carries a uniform linear charge

Serga [27]

Answer:

Explanation:

Let the plastic rod extends from - L to + L .

consider a small length of dx on the rod on the positive x axis at distance x . charge on it = λ dx where λ is linear charge density .

It will create a field at point P on y -axis . Distance of point P

= √ x² + .15²

electric field at P due to small charged length

dE = k λ dx x / (x² + .15² )

Its component along Y - axis

= dE cosθ where θ is angle between direction of field dE and y axis

= dE x .15 / √ x² + .15²

= k λ dx .15 / (x² + .15² )³/²

If we consider the same strip along the x axis at the same position on negative x axis , same result will be found . It is to be noted that the component of field in perpendicular to y axis will cancel out each other . Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L

E = ∫ k λ .15 / (x² + .15² )³/² dx

= k λ x L / .15 √( L² / 4 + .15² )

6 0

4 years ago