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3 years ago

The electric resistance in a length of wire is doubled when the wire is _________.

Physics

1 answer:

Serjik [45]3 years ago

3 0

When the wire is twice as long. Also when the cross sectional area is halved

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At what angle two forces P + Q and (P - Q) act so that their resultant is :

stiv31 [10]

Use resultant formula

$\boxed{\sf R=\sqrt{A^2+B^2+2ABcos\theta}}$

A be p+q and B be p-q

$\\ \rm\Rrightarrow R=\sqrt{3p^2+q^2}$

$\\ \rm\Rrightarrow \sqrt{(p+q)^2+(p-q)^2+2(p+q)(p-q)cos\alpha}=\sqrt{3p^2+q^2}$

$\\ \rm\Rrightarrow 2p^2+2q^2+2(p^2-q^2)cos\alpha=3p^2+q^2$

$\\ \rm\Rrightarrow 2cos\alpha=1$

$\\ \rm\Rrightarrow cos\alpha=\dfrac{1}{2}$

$\\ \rm\Rrightarrow \alpha=\dfrac{\pi}{3}$

$\\ \rm\Rrightarrow 2(p^2+q^2)+2(p^2-q^2)cos\beta=2(p^2+q^2)$

$\\ \rm\Rrightarrow 2cos\beta=0$

$\\ \rm\Rrightarrow cos\beta=0$

$\\ \rm\Rrightarrow \beta=\dfrac{\pi}{2}$

$\\ \rm\Rrightarrow 2(p^2+q^2)+2(p^2-q^2)cos\gamma=p^2+q^2$

$\\ \rm\Rrightarrow 2(p^2-q^2)cos\gamma=-(p^2+q^2)$

$\\ \rm\Rrightarrow cos\gamma =\dfrac{q^2-p^2}{2(p^2-q^2)}$

$\\ \rm\Rrightarrow \gamma=cos^{-1}\left(\dfrac{q^2-p^2}{2(p^2+q^2)}\right)$

8 0

2 years ago

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What should be the spring constant k of a spring designed to bring a 1200 kg car to rest from a speed of 85 km/h so that the occ

borishaifa [10]

Answer:

k = 5178.8 N/m

Explanation:

As we know that spring mass system will oscillate at angular frequency given as

$\omega = \sqrt{\frac{k}{m}}$

now we have

$\omega = \sqrt{\frac{k}{1200}}$

now the maximum acceleration of the spring block system is at its maximum compression state which is given as

$a = \omega^2 A$

here A= maximum compression of the spring

so here in order to find maximum compression of the spring we will use energy conservation as we know that initial total kinetic energy of the car will convert into spring potential energy

$\frac{1}{2}mv^2 = \frac{1}{2}kA^2$

here we know that

v = 85 km/h

$v = 85 \times\frac{1000}{3600} = 23.61 m/s$

now we have

$(1200)(23.61^2) = kA^2$

$A^2 = \frac{6.68 \times 10^5}{k}$

now from above equation of acceleration we have

$5.0 g = (\frac{k}{m})\sqrt{\frac{6.68 \times 10^5}{k}}$

$5.0(9.81) = \sqrt{k}(0.68)$

$k = 5178.8 N/m$

6 0

3 years ago

The downsprue leading into the runner of a certain mold has a length of 175 mm. The cross-sectional area at the base of the spru

adoni [48]

Answer:

(a) Velocity at bottom is 1.85 m/s

(b) Volume flow rate is 7.4 x 10⁻⁴ m³/s.

Explanation:

(a)

Applying Bernoulli's Equation on both ends of the down sprue, with the assumptions that every point is at atmospheric pressure and the liquid metal at the pouring basin is at zero velocity. The equation then becomes:

V = √2gh

where,

V = velocity at bottom of down sprue

h = height of down sprue = 175 mm = 0.175 m

V = √2(9.8 m/s²)(0.175 m)

V = 1.85 m/s

(b)

The volume flow rate is given as:

Volume Flow Rate = (V)(A)

where,

V = velocity at bottom = 1.85 m/s

A = Area of bottom = 400 mm² = 0.0004 m²

Therefore,

Volume Flow Rate = (1.85 m/s)(0.0004 m²)

Volume Flow Rate = 7.4 x 10⁻⁴ m³/s = 740 cm³/s

(c)

The time required to fill the cavity is given as:

Volume Flow Rate = V/t

where,

V = Volume of mold Cavity = 0.001 m³

t = time required to fill the cavity = ?

Therefore,

t = V/Volume Flow Rate

t = 0.001 m³/7.4 x 10⁻⁴ m³/s

t = 1.35 s

5 0

3 years ago

Canola oil is less dense than water, so it floats on water, but its index of refraction is 1.47, higher than that of water. When

kupik [55]

Answer:

therefore critical angle c= 69.79°

Explanation:

Canola oil is less dense than water, so it floats over water.

Given $n_{canola}= 1.47$

which is higher than that of water

refractive index of water $n_{water}=1.33$

to calculate critical angle of light going from the oil into water

we know that

$sinc= \frac{n_{water}}{n_{canola}}$

now putting values we get

$sinc= \frac{1.33}{1.47}$

c= $sin^{-1}(\frac{1.33}{1.47} )$

c=69.79°

therefore critical angle c= 69.79°

8 0

3 years ago

A 10.0 cm object is 5.0 cm from a concave mirror that has a focal length of 12 cm. What is the distance between the image and th

fiasKO [112]

Let's use the mirror equation to solve the problem:
$\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i}$
where f is the focal length of the mirror, $d_o$ the distance of the object from the mirror, and $d_i$ the distance of the image from the mirror.
For a concave mirror, for the sign convention f is considered to be positive. So we can solve the equation for $d_i$ by using the numbers given in the text of the problem:
$\frac{1}{12 cm}= \frac{1}{5 cm}+ \frac{1}{d_i}$
$\frac{1}{d_i}= -\frac{7}{60 cm}$
$d_i = -8.6 cm$
Where the negative sign means that the image is virtual, so it is located behind the mirror, at 8.6 cm from the center of the mirror.

6 0

3 years ago

Read 2 more answers