Tony ran 600 meters in 60 seconds. What was Tony's speed during the<br> race?

A conductor is formed into a loop that encloses an area of 1m^2. The loop is oriented at a angle 30 degree with the xy plane. A

choli [55]

Answer:

$\frac{dB}{dt} = 11.55 T/s$

Explanation:

As we know that flux of magnetic field from the closed loop is given by

$\phi = BAcos30[tex]now by Faraday's law we know[tex]EMF = \frac{d\phi}{dt}$

now we will have

$EMF = Acos30\frac{dB}{dt}$

now we have

$A = 1 m^2$

EMF = 10 Volts

$10 = 1 (cos30)\frac{dB}{dt}$

$\frac{dB}{dt} = \frac{20}{\sqrt3}$

$\frac{dB}{dt} = 11.55 T/s$

5 0

3 years ago

A particle oscillates harmonically x = A cos(ωt + φ0), with amplitude 9 m, angular frequency π s −1, and initial phase π 3 radia

Gekata [30.6K]

Answer:

$t = \frac{5}{12} s$

Explanation:

As we know that the equation of particle position is given as

$x = A cos(\omega t + \phi_0)$

Now the speed of the particle is given as

$v = A\omega sin(\omega t + \phi_0)$

now we know that potential energy and kinetic energy is equal

so we have

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

so we will have

$A^2\omega^2 sin^2(\omega t + \phi_0) = A^2\omega^2 cos^2(\omega t + \phi_0)$

$tan^2(\omega t + \phi_0) = 1$

$\omega t + \phi_0 = \frac{\pi}{4} or \frac{3\pi}{4}$

$\pi t = \frac{3\pi}{4} - \frac{\pi}{3}$

$\pi t = \frac{5\pi}{12}$

$t = \frac{5}{12} s$

6 0

3 years ago

By means of a rope whose mass is negligible, two blocks are suspended over a pulley, as the drawing shows. The pulley can be tre

tankabanditka [31]

Answer:

Mass of the pull is 77 kg

Explanation:

Here we have for

Since the rope moves along with pulley, we have

For the first block we have

T₁ - m₁g = -m₁a = -m₁g/4

T₁ = 3/4(m₁g) = 323.4 N

Similarly, as the acceleration of the second block is the same as the first block but in opposite direction, we have

T₂ - m₂g = m₂a = m₂g/4

T₂ = 5/4(m₂g) = 134.75 N

T₂r - T₁r = I·∝ = 0.5·M·r²(-α/r)

∴ $M = -\frac{2}{a} (T_2-T_1)$

$M = -\frac{2}{2.45} (134.75-323.4) = 77 \, kg$

Mass of the pull = 77 kg.

5 0

4 years ago

A permanent magnet creates a magnetic field at the origin with strength Bperm-1T. A current-carrying wire is oriented such that

bazaltina [42]

Answer:

Part a)

$B_{net} = (1 + 4 \times 10^{-6})T$

Part b)

$B_{net} = (1 - 4 \times 10^{-6})T$

Explanation:

Part a)

Since the two magnetic field is in same direction

so the net magnetic field is algebraic sum of magnetic field due to both

so here magnetic field of wire is given as

$B = \frac{\mu_0 i}{2\pi r}$

here we know that

I = 2 A

r = 5 cm

so we will have

$B = \frac{2 \times 10^{-7} (2)}{0.05}$

$B = 4 \times 10^{-6} T$

So net magnetic field is given as

$B_{net} = (1 + 4 \times 10^{-6})T$

Part b)

When direction of current is reversed then the direction of magnetic field is also reversed

So we will have

$B_{net} = (1 - 4 \times 10^{-6})T$

8 0

3 years ago

A cubical box, 5.00 cm on each side, is immersed in a fluid. The gauge pressure at the top surface of the box is 594 Pa and the

Zolol [24]

Answer:

The density of the fluid is 1100 kg/m³.

Explanation:

Given that,

Height = 5.00 cm

Pressure at top =594 Pa

Pressure at bottom = 1133 Pa

We need to calculate the change in pressure

Using formula of change in pressure

$\Delta P=P_{b}-P_{t}$

Where, $P_{b}$ = Pressure at bottom

$P_{t}$ = Pressure at top

put the value into the formula

$\Delta P=1133-594$

$\Delta P=539\ Pa$

Using formula of pressure for density

$\Delta P = \rho g h$

$\rho =\dfrac{\Delta P}{gh}$

Where, $\rho$ = density

P = pressure

h = height

Put the value in to the formula

$\rho =\dfrac{539}{5.00\times10^{-2}\times9.8}$

$\rho =1100\ kg/m^3$

Hence, The density of the fluid is 1100 kg/m³.

4 0

4 years ago

Tony ran 600 meters in 60 seconds. What was Tony's speed during the race?