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ser-zykov [4K]
2 years ago
11

A toy car has a battery-powered fan attached to it such that the fan creates a constant force that is exerted on the car so that

it is propelled in the opposite direction in which the fan blows air. The car has a carriage that allows a student to attach objects of different masses, as shown above. The fan has only one speed setting. All frictional forces are considered to be negligible. Which of the following procedures could be used to determine how the mass of the fan-car-object system affects the acceleration of the system?
Physics
2 answers:
Andreas93 [3]2 years ago
7 0

Answer:

Measure the mass of the system using a balance, activate the fan, measure the distance traveled by the system at a known time by using a stopwatch, and repeat the experiment for several trials with different objects added to the carriage.

Explanation:

swat322 years ago
4 0

Procedures <em><u>point C</u></em> could be used to determine how the mass of the fan-car-object system affects the acceleration of the system

<h3>Further explanation </h3>

Newton's 2nd law explains that the acceleration produced by the resultant force on an object is proportional and in line with the resultant force and inversely proportional to the mass of the object

<h3>∑F = m. a </h3>

\large{\boxed{\bold{a=\frac{\sum F}{m}}}

F = force, N

m = mass = kg

a = acceleration due to gravity, m / s²

we complete the available answer choices

A).Measure the mass of the system using a balance, activate the fan, measure the distance traveled by the system at a known time by using a stopwatch, and repeat the experiment for several trials with different objects added to the carriage.

B) Measure the mass of the system using a balance, activate the fan, use a meterstick and stopwatch to measure the initial and final speeds of the system, and repeat the experiment for several trials with different objects added to the carriage.

C)Measure the mass of the system using a balance, connect a spring scale to the back of the car, measure the amount of force required to hold the system at rest, and repeat the experiment for several trials with different objects added to the carriage.

D)Measure the mass of the system using a balance, activate the fan, use a stopwatch to record the time it takes for the system to travel before the battery of the fan no longer works, and repeat the experiment for several trials with different objects added to the carriage.

From the choices above, we choose point C because to find out the effect of mass on acceleration, we need the force value (F) from the spring scale and the mass of objects of different masses using a balance

If we look at the formula, we can analyze and estimate that the greater the mass given, the smaller the system acceleration

<h3>Learn more</h3>

law of motion  

brainly.com/question/75210  

displacement of A skateboarder

brainly.com/question/1581159

The distance of the elevator  

brainly.com/question/8729508  

Keywords : system acceleration, A toy car, fan

#LearnWithBrainly

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padilas [110]

Answer:

 a)    λ = 189.43 10⁻⁹ m  b)    λ = 269.19 10⁻⁹ m

Explanation:

The diffraction network is described by the expression

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Where m corresponds to the diffraction order

Let's use trigonometry to find the breast

        tan θ = y / L

The diffraction spectrum is measured at very small angles, therefore

      tan θ = sin θ / cos θ = sin θ

We replace

      d y / L = m λ

Let's place in the first order m = 1

Let's look for the separation of the lines (d)

     d = λ  L / y

     d = 501 10⁻⁹ 9.95 10⁻² / 15 10⁻²

     d = 332.33 10⁻⁹ m

Now we can look for the wavelength of the other line

     λ  = d y / L

    λ  = 332.33 10⁻⁹ 8.55 10⁻²/15 10⁻²

    λ = 189.43 10⁻⁹ m

Part B

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      λ  = 332.33 10⁻⁹ 12.15 10⁻² / 15 10⁻²

      λ = 269.19 10⁻⁹ m

4 0
2 years ago
In a 1.25-T magnetic field directed vertically upward, a particle having a charge of magnitude 8.50μC and initially moving north
Goshia [24]

Answer:

a) The sign of the charge is positive.

b) The magnetic force on the particle is 0.050 newtons.

Explanation:

The magnetic force F on a moving charge with velocity v passing through a magnetic field B is:

\overrightarrow{F}=q\overrightarrow{v}\times\overrightarrow{B}(1)

a)

Because it is a cross product, we can find the direction of the force using the right-hand rule, that is too the direction of the movement. We have two possibilities here because the velocity vector and magnetic field are perpendicular: the particle deflects towards east or toward west, which depends on the charge of the particle. Note that if you put your right hand fingers, except thumb, pointing towards north (direction of velocity) and later close them in the direction of the magnetic field, if you maintain your thumb perpendicular to this movement it will point towards east (See figure), so that will be de direction of the force if the charge is positive, but if the charge is negative, the direction will be opposite (towards west). So the charge has to be positive to deflects towards east.

b)

Now by 1:

F=qvB\sin\theta=(8.50\times10^{-6})(4750)(1.25)\sin90\simeq\mathbf{0.050\,N}

5 0
2 years ago
A thin, horizontal, 18-cm-diameter copper plate is charged to -3.8 nC. Assume that the electrons are uniformly distributed on th
son4ous [18]

Answer:

Part a)

E = 8436.7 N/C

Part b)

E = 8436.7 N/C

Explanation:

Part a)

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\sigma = \frac{Q}{A}

Q = -3.8 nC

A = \pi(0.09)^2

A = 0.025 m^2

\sigma = \frac{-3.8\times 10^{-9}}{0.025}

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now the electric field is given as

E = \frac{-1.5 \times 10^{-7}}{2(8.85 \times 10^{-12})}

E = 8436.7 N/C

Part b)

Now since the electric field is required at same distance on other side

so the field will remain same on other side of the plate

E = 8436.7 N/C

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2 years ago
1. A 9.4×1021 kg moon orbits a distant planet in a circular orbit of radius 1.5×108 m. It experiences a 1.1×1019 N gravitational
sattari [20]

Answer:

26 days

Explanation:

m = 9.4×1021 kg

r= 1.5×108 m

F = 1.1×10^ 19 N

We know Fc = \frac{m v^{2} }{r}

==> 1.1 × 10^{19} = (9.4 × 10^{21} × v^{2} ) ÷ 1.5 × 10^{8}

==> 1.1 × 10^{19} = v^{2} × 6.26×10^{13}

==> v^{2} =  1.1 × 10^{19} ÷ 6.26×10^{13}

==> v^{2} = 0.17571885 × 10^{6}

==> v= 0.419188323 × 10^{3} m/sec

==> v= 419.188322834 m/s

Putting value of r and v from above in ;

T= 2πr ÷ v

==> T= 2×3.14×1.5×10^{8} ÷ 0.419188323 × 10^{3}

==> T = 22.472× 100000 = 2247200 sec

but

86400 sec = 1 day

==> 2247200 sec= 2247200 ÷ 86400 = 26 days

3 0
2 years ago
Read 2 more answers
A block spring system oscillates on a frictionless surface with an amplitude of 10\text{ cm}10 cm and has an energy of 2.5 \text
antoniya [11.8K]

Answer:

The energy of the system is 15 J.

Explanation:

Given that,

Energy E = 2.5 J

Amplitude = 10 cm

We need to calculate the spring constant

Using formula of mechanical energy of the system

E=\dfrac{1}{2}kA^2

Put the value into the formula

2.5=\dfrac{1}{2}k\times(10\times10^{-2})^2

k=\dfrac{2.5\times2}{(10\times10^{-2})^2}

k=500\ N/m

If the block is replaced by a block with twice the mass of the original block

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We need to calculate the energy

Using formula of mechanical energy

E=\dfrac{1}{2}kA^2

Put the value into the formula

E=\dfrac{1}{2}\times500\times(6\times10^{-2})

E=15\ J

Hence, The energy of the system is 15 J.

8 0
2 years ago
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