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4 years ago

A strong inductive argument must have true premises True False

Physics

1 answer:

Amiraneli [1.4K]4 years ago

6 0

That is true imo not false

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Please answer these!

solong [7]

6. f = 128.62 $s^{-1}$ , T= 0.0077775 s

7. f = 2.2 * $10^{4}$ $s^{-1}$ , T = 4.545 * $10^{-5}$ s

8. 32.64 $s^{-1}$

9. 3.29 * $10^{14}$ $s^{-1}$

Explanation:

Step 1:

For light and sound v = fλ

where v represents the velocity

f represents the frequency

λ represents the wavelength

λ = 2.69 m

v = 346 m/s

f = v/λ = 346/2.69 = 128.62 $s^{-1}$

Time period is the reciprocal of frequency

T = 1/128.62 = 0.0077775 s

Step 2:

λ = 110 cm = 1.1 m

v = 2.42* $10^{4}$ m/s

f = 2.42* $10^{4}$ /1.1 = 2.2 * $10^{4}$ $s^{-1}$

T = 1/(2.2* $10^{4}$ ) = 4.545 * $10^{-5}$ s

Step 3:

λ = 10.6 m

v = 346 m/s

f = v/λ = 346/10.6 = 32.64 $s^{-1}$

Step 4:

λ = 5.89 * $10^{-7}$ m

v = 1.94 * $10^{8}$ m/s

f = v/λ = 3.29 * $10^{14}$ $s^{-1}$

8 0

4 years ago

Problem 21-40a:

Greeley [361]

Answer:

Part a)

$\frac{F_e}{F_g} = 2.74 \times 10^{12}$

Part b)

$q = 3.37 \times 10^{-4} C$

Explanation:

As we know that electric force on electric charge is given as

$F = qE$

here we have

$q = 1.6 \times 10^{-19}C$

E = 153 N/C

now force is given as

$F = (1.6 \times 10^{-19})(153) = 2.45 \times 10^{-17} N$

Gravitational force on electric charge near surface of earth is given as

$F_g = mg$

$F_g = (9.1 \times 10^{-31})(9.81) = 8.93 \times 10^[-30} N$

now the ratio of two forces is given as

$\frac{F_e}{F_g} = \frac{2.45 \times 10^{-17}}{8.93 \times 10^{-30}}$

$\frac{F_e}{F_g} = 2.74 \times 10^{12}$

Part b)

Now the ball is balanced by the electric force and the force of gravity on it

so here we have

$F_g = qE$

$mg = qE$

$(5.25 \times 10^{-3})(9.81) = q(153)$

here we have

$q = 3.37 \times 10^{-4} C$

8 0

4 years ago

A light ray incident from medium 1 to medium 2, where n1>n2. When the incident angle exceed the critical angle ac, the refrac

vovikov84 [41]

Explanation:

(a)

Critical angle is the angle at the angle of refraction is 90°. After the critical angle, no refraction takes place.

Using Snell's law as:

$n_1\times {sin\theta_i}={n_2}\times{sin\theta_r}$

Where,

${\theta_i}$ is the angle of incidence

${\theta_r}$ is the angle of refraction = 90°

${n_2}$ is the refractive index of the refraction medium

${n_1}$ is the refractive index of the incidence medium

Thus,

$n_1\times {sin\ \theta_{critical}}={n_2}\times{sin\ 90^0}$

The formula for the calculation of critical angle is:

${sin\theta_{critical}}=\frac {n_2}{n_1}$

Where,

${\theta_{critical}}$ is the critical angle

(b)

No it cannot occur. It only occur when the light ray bends away from the normal which means that when it travels from denser to rarer medium.

7 0

4 years ago

A circular conducting loop with a radius of 0.10 m and a small gap filled with a 10.0 ȍresistor is oriented in the xy-plane. If

lidiya [134]

Answer:

The magnitude of the current is 5.45 mA.

Explanation:

Given that,

Resistance = 10.0 ohm

Radius = 0.10 m

Magnetic field = 1.0 T

Angle = 30°

Increase magnetic field = 7.0 T

Time t = 3.0 s

Number of turns = 1

We need to calculate the initial flux

Using formula of flux

$\phi=NB_{1}A\cos\theta$

Put the value into the formula

$\phi=1\times1.0\times\pi\times(0.10)^2\times\cos30^{\circ}$

$\phi=1\times1.0\times\pi\times(0.10)^2\times\dfrac{\sqrt{3}}{2}$

$\phi=0.027\ wb$

We need to calculate the final flux

$\phi=1\times7.0\times\pi\times(0.10)^2\times\dfrac{\sqrt{3}}{2}$

$\phi=0.1904\ wb$

We need to calculate the induced emf

Using formula of emf

$\epsilon=\dfrac{\phi_{f}-\phi_{i}}{t}$

Put the value into the formula

$\epsilon=\dfrac{0.1904-0.027}{3.0}$

$\epsilon=0.0545\ V$

We need to calculate the current

Using formula of current

$I=\dfrac{\epsilon}{R}$

Put the value into the formula

$I=\dfrac{0.0545}{10.0}$

$I=5.45\ mA$

Hence, The magnitude of the current is 5.45 mA.

4 0

3 years ago

An arrow of 43 g moving at 84 m/s to the right, strikes an apple at rest. The arrow sticks to the apple and both travel at 16.8

Aloiza [94]

Answer:

<em>The mass of the apple is 0.172 kg (172 g)</em>

Explanation:

<u>The Law Of Conservation Of Linear Momentum </u>

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is

P=mv.

If we have a system of two bodies, then the total momentum is the sum of both momentums:

$P=m_1v_1+m_2v_2$

If a collision occurs and the velocities change to v', the final momentum is:

$P'=m_1v'_1+m_2v'_2$

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2$

If both masses stick together after the collision at a common speed v', then:

$m_1v_1+m_2v_2=(m_1+m_2)v'$

We are given the mass of an arrow m1=43 g = 0.043 kg traveling at v1=84 m/s to the right (positive direction). It strikes an apple of unknown mass m2 originally at rest (v2=0). The common speed after they collide is v'=16.8 m/s.

We need to solve the last equation for m2:

$m_2v_2-m_2v'=m_1v'-m_1v_1$