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scZoUnD [109]
3 years ago
12

A car is moving at a velocity of 25 km/h and increases velocity to 1200 km/h in 2 min. what is the acceleration?

Physics
1 answer:
OlgaM077 [116]3 years ago
3 0

Answer:

a = 2.72 [m/s2]

Explanation:

To solve this problem we must use the following kinematics equation:

v_{f} =v_{o} + a*t

where:

Vf = final velocity = 1200 [km/h]

Vo = initial velocity = 25 [km/h]

t = time = 2 [min] = 2/60 = 0.0333 [h]

1200 = 25 + (a*0.0333)

a = 35250.35 [km/h2]

if we convert these units to units of meters per second squared

35250.35[\frac{km}{h^{2} }]*(\frac{1}{3600^{2} })*[\frac{h^{2} }{s^{2} } ]*(\frac{1000}{1} )*[\frac{m}{km} ] = 2.72 [\frac{m}{s^{2} } ]

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The drawing shows a plot of the output emf of a generator as a function of time t. The coil of this device has a cross-sectional
adell [148]

This question is incomplete, the missing image uploaded along this answer below;

Answer:

a) frequency is 2.381 Hz

b) the angular frequency/speed is 14.96 rad/s

c) the magnitude of the magnetic field is 0.5199 T

Explanation:

Given that;

cross sectional Area A = 0.018 m²

Number of turns N = 200

from the diagram. maximum time T = 0.42 sec

a) the frequency f of the generator in hertz

frequency = 1 / T

we substitute

frequency = 1 / 0.42 = 2.381 Hz

Therefore, frequency is 2.381 Hz

b)  the angular frequency in rad/s

angular speed ω = 2πf

we substitute

ω = 2π × 2.381

ω = 14.96 rad/s

Therefore, the angular frequency/speed is 14.96 rad/s

c) the magnitude of the magnetic field.

to determine the magnitude of the magnetic field, we use the following expression;

e = NBAω

from the diagram, e = 28.0 V

so we substitute

28.0 V = 200 × B × 0.018 × 14.96

28 = 53.856B

B = 28 / 53.856

B = 0.5199 T

Therefore, the magnitude of the magnetic field is 0.5199 T

6 0
3 years ago
11. A fundamental property of light is that it: 15
myrzilka [38]

Answer:

Curves around objects

Explanation:

Diffraction is a property of light described by bending of light around an object. This ability of light to bend around edges has facilitated optical effects of light where there is interference of light waves. Other properties of light are: reflection, refraction, polarization, scattering of light, and interference of light.

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3 years ago
Followed by the previous question: presume that the electron performs a uniform circular motion around the hydrogen nucleus. Wha
Ksivusya [100]

Answer:

A_c=87.73*10^{21}m/s

Explanation:

From the question we are told that

r=5\times 10^{-11}

T=1.5 \times 10^{-16}

Generally the equation for velocity is mathematically given as

Velocity (v)=\frac{2 \pi r}{t}

V=\frac{2 \pi (5*10^{-11})}{1.5*10^{-16}}

V=\frac{2 \pi (5*10^{-11})}{1.5*10^{-16}}

Generally the equation for Centripetal acceleration is mathematically given as

A_c=\frac{V^2}{r}

A_c=(\frac{20.944*10^5)}{r5*10^{-11}}

A_c=87.73*10^{21}m/s

8 0
2 years ago
Which of the following statements about the resistsance of a wire is correct? Select THREE answers.
viva [34]

Answer:

Its a

Explanation:

4 0
3 years ago
03: A mass with a 60 g vibrate at the end of a spring. The amplitude of the motion is 0.394 ft
Flauer [41]

Answer:

a) T = 1.69 s, b)  k = 0.825 N / m, c)  v = 1.46 feet/s, d) a = 5.41 ft / s²,

e)   v = - 1,319 ft / s,    a = - 2.70 ft / s², f) K = 4.8 10⁻³ J, U = 1.49 10⁻³ J

Explanation:

In a mass-spring system with simple harmonic motion, the angular velocity is

         w = \sqrt{\frac{k}{m} }

a) find the period

angular velocity, frequency, and period are related

         w = 2π f = 2π / T

          f = 1 / T

          T = 1 / f

           T = 1 / 0.59

           T = 1.69 s

b) the spring constant

         w = 2π f

         w = 2π 0.59

         w = 3.70 rad / s

         w² = k / m

          k = w² m

          k = 3.70² 0.060

          k = 0.825 N / m

c) the maximum speed

simple harmonic movement is described by the expression

          x = A cos (wt + Ф)

speed is defined by

         v =\frac{dx}{dt}

          v = -A w sin (wt + fi)

the speed is maximum when the cosine is ± 1

          v = A w

          v = 0.394 3.70

          v = 1.46 feet/s

d) maximum acceleration

            a = \frac{dv}{dt}

            a = - A w² cos wt + fi

the acceleration is maximum when the cosine is ±1

            a = A w²

            a = 0.394 3.70²

            a = 5.41 ft / s²

e) velocity and acceleration for x = 6 cm

let's reduce the cm to feet

            x = 6 cm (1 foot / 30.48 cm) = 0.1969 foot

Before doing this part we must find the phase angle (Ф), the most common way to start the movement is to move the spring a small distance and release it, so its initial speed is zero for t = 0 s

let's use the expression for the velocity

           v = -A w sin (0 + Фi)

           0 = - A w sin Ф

so sin Ф = 0 which implies that Фi = 0

the equation of motion is

            x = A cos wt

            x = 0.394 cos 3.70t

we substitute

           0.1969 = 0.394 cos 370t

           3.70 t = cos⁻¹ (0.1969 / 0.394)

let's not forget that the angle is in radians

           3.70, t = 1.047

           t = 1.047 / 3.70

           t = 0.2826 s

we substitute this time in the equation for velocity and acceleration

           v = - Aw sin wt

           v = - 0.394 3.70 sin 3.70 0.2826

           v = - 1,319 ft / s

           a = - A w² cos wt

           a = - 0.394 3.70² cos 3.70 0.2826

           a = - 2.70 ft / s²

f) the kinetic and potential energy at this point

           K = ½ m v²

let's slow down to the SI system

           v = 1.319 ft / s (1 m / 3.28 ft) = 0.402 m / s

           

           K = ½ 0.060 0.402²

           K = 4.8 10⁻³ J

           U = ½ k x²

           U = ½ 0.825 0.06²

           U = 1.49 10⁻³ J

5 0
3 years ago
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