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hoa [83]
4 years ago
11

Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Expl

ain

Physics
1 answer:
professor190 [17]4 years ago
3 0

Complete question:

A force F is applied to the block as shown (check attached image). With an applied force of 1.5 N, the block moves with a constant velocity.

Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Explain

Answer:

The applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N

Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the initially applied force.

Explanation:

Given;

magnitude of applied force, F = 1.5 N

Apply Newton's second law of motion;

F = ma

F = m(\frac{v}{t} )\\\\F = \frac{m}{t} v\\\\Let \ \frac{m}{t} \ be \ constant = k\\F = kv\\\\k = \frac{F}{v} \\\\\frac{F_1}{v_1} = \frac{F_2}{v_2}

The applied force needed to keep the box moving with a constant velocity that is twice as fast as before;

\frac{F_1}{v_1} = \frac{F_2}{v_2} \\\\(v_2 = 2v_1, \ and \ F_1 = 1.5N)\\\\\frac{1.5}{v_1} = \frac{F_2}{2v_1} \\\\1.5 = \frac{F_2}{2}\\\\F_2 = 2*1.5\\\\F_2 = 3 N

Therefore, the applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N

Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the applied force.

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Answer : F_{x} = 45.6\times10^{-2}\ N, F_{y} = 3040\times10^{-5} N and F_{z} = 0\ N

Explanation :

Given that,

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Electric field in x-direction E_{x} = 5.70\times10^{3}\ N/C

Electric field in y- direction E_{y} = 380\ N/C

Electric field in z - direction E_{z} = 0

Now, using formula

F = q E

Now, the force on the object in x- direction

F_{x} = 8.00\times10^{-5} C \times5.70\times10^{3} N/C

F_{x} = 45.6\times10^{-2}\ N

The force on the object in y- direction

F_{y} = 8.00\times10^{-5}\ C\times 380\ N/C

F_{y} = 3040\times10^{-5} N

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F_{z} = 0\ N

Hence, this is the required solution.




8 0
4 years ago
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.55 times a second. A tack is stuck in the tire a
earnstyle [38]

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Tangential speed = 5.72 m/s

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Explanation:

The tangential speed, V, is given by

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where \omega is the angular speed and is given by 2\pi f (f is the angular frequency or frequency of rotation)

Thus,

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The centripetal acceleration,a, is given by

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7 0
4 years ago
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Answer:

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We have given that the there is parallel circuit with three different branches.

Current in each branch is I_1=2.8A, I_2=4.1A,I_2=4.8A

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It is known that in parallel circuit current is sum of current in each branches

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I=I_1+I_2+I_3

I=2.8+4.1+4.8=11.7A

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8 0
3 years ago
If the torque required to loosen a nut that holds a wheel on a car has a magnitude of 55 n·m, what force must be exerted at the
erastova [34]

Either 175 N or 157 N depending upon how the value of 48° was measured from.    
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Since we need a torque of 55 N·m to loosen the nut and our lug wrench is 0.47 m long, that means that we need 55 N·m / 0.47 m = 117 N of usefully applied force in order to loosen the nut. This figure will be used for both possible angles.    
Ideally, the force will have a 0° degree difference from the normal and 100% of the force will be usefully applied. Any value greater than 0° will have the exerted force reduced by the cosine of the angle from the normal. Hence the term "cosine loss".     
If the angle of 48° is from the normal to the lug wrench, the usefully applied power will be:  
U = F*cos(48)  
where  
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So solving for F and calculating gives:  
U = F*cos(48)  
U/cos(48) = F  
117 N/0.669130606 = F  
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So 175 Newtons of force is required in this situation.    
If the 48° is from the lug wrench itself, that means that the force is 90° - 48° = 42° from the normal. So doing the calculation again (this time from where we started plugging in values) we get  
U/cos(42) = F  
117/0.743144825 = F  
157.4390294 = F    
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3 years ago
A block weighing 3.7 kg is suspended from the ceiling of a truck trailer by a hanging bungee cord. The cord has a cross-sectiona
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\Delta L = 10 cm

now we have strain in the string is given as

strain = \frac{\Delta L}{L}

strain = \frac{{10 cm}{43 cm}

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similarly we will have cross-sectional area of the string is given as

A = 40 \times 10^{-6} m^2

now the stress in the string is given as

Stress = \frac{T}{A}

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Stress = \frac{3.7 \times 9.81}{40 \times 10^{-6}}

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Now Young's Modulus is given as

Y = \frac{stress}{strain}

Y = \frac{9.07 \times 10^5}{40\times 10^{-6}}

Y = 2.27 \times 10^{10} N/m^2

5 0
3 years ago
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