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3 years ago

The acceleration of an object would increase if there was an increase in the

Physics

1 answer:

xxMikexx [17]3 years ago

4 0

As per Newton's II law we know that

$F_{net} = ma$

here we know that

$F_{net} = F_{ap} - F_f$

so here we will have

$a = \frac{F_{ap} - F_f}{m}$

so here if we need to increase the acceleration we need to increase the applied force while on increasing the mass or on increasing the friction force the acceleration will decrease.

So here correct answer will be

A) force on the object.

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A particle in a 799=m-long linear particle accelerator is moving at 0.875C. How long does the particle accelerator appear to the

barxatty [35]

the answer is 386m because m = pi mc=2.23

6 0

1 year ago

Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 24 kg and the larger bottom crate has a m

marusya05 [52]

Answer:

The sum of all forces for the two objects with force of friction F and tension T are:

(i) m₁a₁ = F

(ii) m₂a₂ = T - F

1) no sliding infers: a₁ = a₂= a

The two equations become:

m₂a = T - m₁a

Solving for a:

a = T / (m₁+m₂) = 2.1 m/s²

2) Using equation(i):

F = m₁a = 51.1 N

3) The maximum friction is given by:

F = μsm₁g

Using equation(i) to find a₁ = a₂ = a:

a₁ = μs*g

Using equation(ii)

T = m₁μsg + m₂μsg = (m₁ + m₂)μsg = 851.6 N

4) The kinetic friction is given by: F = μkm₁g

Using equation (i) and the kinetic friction:

a₁ = μkg = 6.1 m/s²

5) Using equation(ii) and the kinetic friction:

m₂a₂ = T - μkm₁g

a₂ = (T - μkm₁g)/m₂ = 12.1 m/s²

4 0

3 years ago

Please answer the following question

aleksklad [387]

Answer:

1. Light Bulb

2. A switch

3. A battery

4. Ammeter

5. Resistor

6. Voltmeter

4 0

2 years ago

An object with a mass of 10 kg is rolled down a frictionless ramp from a height of 3 meters. If a factory worker at the bottom o

ruslelena [56]

Answer:

The amount of work the factory worker must to stop the rolling ramp is 294 joules

Explanation:

The object rolling down the frictionless ramp has the following parameters;

The mass of the object = 10 kg

The height from which the object is rolled = 3 meters

The work done by the factory worker to stop the rolling ramp = The initial potential energy, P.E., of the ramp

Where;

The potential energy P.E. = m × g × h

m = The mass of the ramp = 10 kg

g = The acceleration due to gravity = 9.8 m/s²

h = The height from which the object rolls down = 3 m

Therefore, we have;

P.E. = 10 kg × 9.8 m/s² × 3 m = 294 Joules

The work done by the factory worker to stop the rolling ramp = P.E. = 294 joules

8 0

2 years ago

A nonconducting spherical shell, with an inner radius of 4 cm and an outer radius of 6 cm, has charge spread non uniformly throu

Aloiza [94]

In other words a infinitesimal segment dV caries the charge
dQ = ρ dV 

Let dV be a spherical shell between between r and (r + dr): 
dV = (4π/3)·( (r + dr)² - r³ ) 
= (4π/3)·( r³ + 3·r²·dr + 3·r·(dr)² + /dr)³ - r³ ) 
= (4π/3)·( 3·r²·dr + 3·r·(dr)² + /dr)³ ) 
drop higher order terms 
= 4·π·r²·dr 

To get total charge integrate over the whole volume of your object, i.e. 
from ri to ra: 
Q = ∫ dQ = ∫ ρ dV 
= ∫ri→ra { (b/r)·4·π·r² } dr 
= ∫ri→ra { 4·π·b·r } dr 
= 2·π·b·( ra² - ri² ) 

With given parameters: 
Q = 2·π · 3µC/m²·( (6cm)² - (4cm)² ) 
= 2·π · 3×10⁻⁶C/m²·( (6×10⁻²m)² - (4×10⁻²m)² ) 
= 3.77×10⁻⁸C 
= 37.7nC

6 0

3 years ago