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3 years ago

12. A frain moves from rest to a speed of 25 m/s in 30.0 seconds. What is its acceleration?

Physics

1 answer:

ziro4ka [17]3 years ago

6 0

initial velocity=u=0m/s
Final velocity=v=25m/s
Time=t=30s

$\\ \tt\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \tt\longmapsto Acceleration=\dfrac{25-0}{30}$

$\\ \tt\longmapsto Acceleration=\dfrac{25}{30}$

$\\ \tt\longmapsto Acceleration=0.8m/s^2$

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What regulatory requirements must be met prior to a pilot acting as PIC of an aircraft towing a glider?

Klio2033 [76]

Answer:

According to the federal aviation regulation for tow pilot, there are three regulatory requirements to be met.

Explanation:

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4 years ago

Below is a single of DNA, What is the complementary base pair for RNA? AGC, CGT, ATA, GAT

nata0808 [166]

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3 years ago

A cheetah spotted a Gazelle. The cheetah runs at its top speed of 30 m/s for 15 seconds. Durning this time , a gazelle ,160 m fr

Xelga [282]

Answer:

A) 450 m

B) 27 m/s

C) 81 m, 243 m

D) Gazelle

Explanation:

Since, the Cheetah is running at constant speed. Therefore, we use the equation:

s₁ = v₁t

where,

s₁ = distance covered by Cheetah = ?

v₁ = speed of Cheetah = 30 m/s

t = time taken = 15 s

Therefore,

s₁ = (30 m/s)(15 s)

s₁ = 450 m

For final speed of Gazelle at the end of 6 s acceleration time we use 1st equation of motion:

Vf = Vi + at

where,

Vf = Final Speed of Gazelle at the end of 6 s = ?

Vi = Initial Speed of Gazelle = 0 m/s (Since, Gazelle is initially at rest)

t = time taken = 6 s

a = acceleration = 4.5 m/s²

Therefore,

Vf = (0 m/s) + (4.5 m/s²)(6 s)

Vf = 27 m/s

For the distance covered by Gazelle at the end of 6 s acceleration time we use 2nd equation of motion:

s₂ = Vi t + (0.5)at²

where,

Vi = Initial Speed of Gazelle = 0 m/s (Since, Gazelle is initially at rest)

t = time taken = 6 s

a = acceleration = 4.5 m/s²

Therefore,

s₂ = (0 m/s)(6 s) + (0.5)(4.5 m/s²)(6 s)²

s₂ = 81 m

Now, for the distance covered during the last 9 s at constant velocity Vf, we use equation:

s₃ = Vf t

where,

s₃ = distance covered by Gazelle in last 9 s = ?

t = time = 9 s

Therefore:

s₃ = (27 m/s)(9 s)

s₃ = 243 m

We know that, at the end of 15 seconds:

Distance covered by Cheetah = s₁ = 450 m

Distance Covered by Gazelle = s₂ + s₃ = 81 m + 243 m = 324 m

If we take the initial position of Cheetah as origin. Then the positions of both Gazelle and Cheetah with respect to origin will be:

Position of Cheetah = 450 m ahead of origin

Position of Gazelle = 324 m + 160 m = 484 m (Since, Gazelle was initially 160 m ahead of Cheetah)

Hence, it is clear that Gazelle is ahead at the end of 15 s.

5 0

3 years ago

Review. (c) Assume the dipole is a compass needle-a light bar magnet-with a magnetic moment of magnitude μ . It has moment of in

hammer [34]

The frequency of oscillation is $\frac{1}{2\pi } \sqrt{\frac{\mu B}{I} }$ .

<h3>What is a magnetic moment?</h3>

The magnetic moment is the magnetism of a magnet or other item that creates a magnetic field, as well as its orientation and strength. Electromagnets, permanent magnets, elementary particles like electrons, different compounds, and a variety of celestial objects are examples of things that have magnetic moments (such as many planets, some moons, stars, etc). The phrase "magnetic moment" typically refers to a system's magnetic dipole moment, which can be represented by an analogous magnetic dipole, which has a magnetic north and south pole that are barely separated from one another. For sufficiently small magnets or at sufficiently wide distances, the magnetic dipole component is adequate. For extended objects, additional terms may be required in addition to the dipole moment, such as the magnetic quadrupole moment.

$T = \frac{ Id^{2}\theta }{dt^{2} }$

$-\mu B\theta=\frac{ Id^{2}\theta }{dt^{2} }$

$\frac{d^{2}\theta }{dt^{2} } = -(\frac{\mu BI}{\theta} )$

By Comparing the above equation with the SHM equation

$\frac{d^2 \theta} {dt^{2} } = -\omega^{2} \theta$

$\omega^{2} =\frac{ \mu B}{I}$

Frequency = $\frac{\mu}{2\pi }$

= $\frac{\sqrt{\frac{\mu B}{I} } }{2\pi}$

= $\frac{1}{2\pi } \sqrt{\frac{\mu B}{I} }$

To learn more about a magnetic moment, visit:

brainly.com/question/17000031

#SPJ4

8 0

1 year ago

What is the electric force acting between two charges of -0.0050 C and 0.0050 C that are 0.025 m apart?

nika2105 [10]

Answer:

(A) $-3.6\times 10^8\ N$

Explanation:

Given:

Charge of one particle (q₁) = -0.0050 C

Charge of another particle (q₂) = 0.0050 C

Separation between them (d) = 0.025 m

We know that, from Coulomb's law, electric force acting between two charged particles is given as:

$F_e=\dfrac{kq_1q_2}{d^2}\\\\Where,k\to Coulomb's\ constant = 9\times 10^9\ N\cdot m^2/C^2$

Plug in the given values and solve for electric force, $F_e$ . This gives,

$F_e=\frac{(9\times 10^9\ N\cdot m^2/C^2) (-0.0050\ C)(0.0050\ C)}{(0.025\ m)^2}\\\\F_e=\frac{-2.25\times 10^{-4}\times 10^9}{6.25\times 10^{-4}}\ N\\\\F_e=-3.6\times 10^8\ N$

Therefore, option (A) is correct. Negative sign implies that the nature of electric force is attraction.

3 0

3 years ago