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Kruka [31]
3 years ago
10

If the mass of both weights is 225 gm, the first mass is located 20∘ north of east, the second mass is located 20∘ south of east

, and the transducer sensitivity is 0.5 volts/Newton, how large a voltage do you expect to measure? Assume the transducer has been properly zeroed so that V = 0 when F3=0. Please express your answers with 1 decimal place.
Physics
1 answer:
Montano1993 [528]3 years ago
7 0

Answer:

The voltage is 2.114 V.

Explanation:

Given that,

Mass of both weights = 225 gm

Transducer sensitivity = 0.5 V/N

The first mass is located 20∘ north of east, the second mass is located 20∘ south of east,

We need to calculate the net equivalent force

Using formula of force

F_{3}=m_{1}g\cos\theta+m_{2}g\cos\theta

F_{3}=2mg\cos\theta

Put the value into the formula

F_{3}=2\times0.225\times10\cos20^{\circ}

F_{3}=4.228\ N

We need to calculate the voltage

Using formula of voltage

Voltage =sensitivity\times F_{3}

Put the value into the formula

Voltage=0.5\times4.228

Voltage =2.114\ V

Hence, The voltage is 2.114 V.

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A ball is shot from the ground straight up into the air with initial velocity of 42 ft/sec. Assuming that the air resistance can
Volgvan

Answer:

Maximum height of the ball, h(t) = 27.56 m

Explanation:

It is given that, a ball is shot from the ground straight up into the air with initial velocity of 42 ft/sec.      

The height of the ball as a function of time t is given by :

h(t)=h_o+v_ot-16t^2

h₀ is initial height, h₀ = 0

So, h(t)=42t-16t^2 .........(1)

For maximum/minimum height,  \dfrac{dh(t)}{dt}=0

42-32t=0...(2)

t = 1.31 s

Differentiating equation (2) wrt t

h''(t) = -32 < 0

So, at t = 1.31 seconds we will get the maximum height.

Put the value of t in equation (1)

h(t)=42\times 1.31-16\times (1.31)^2

h(t) = 27.56 m

Hence, this is the required solution.

7 0
4 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}

5 0
3 years ago
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LuckyWell [14K]

Answer:

B

Explanation:

The correct answer is B: in the northwest.

6 0
2 years ago
A passenger walks from one side of a ferry to the other as it approaches a dock.Passenger's velocity is 1.55 m/s due north relat
victus00 [196]

Answer:\theta =33.22^{\circ}

Explanation:

Given

Velocity of Passenger w.r.t to Ferry

v_{pf}=v_p-v_f=1.55\hat{j}-------1

Velocity of Passenger w.r.t to water

v_{pw}=v_p-v_w=4.5\left ( -\hat{i}+\hat{j}\right )--------2

Subtract 2 from 1

v_f-v_w=-4.5\hat{i}+2.95\hat{j}

(b)direction

tan\theta =\frac{2.95}{4.5}=0.655

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Part A
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Answer:

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Explanation:

5 0
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