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3 years ago

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If a vehicle is moving at 45m/s initially 36m/s after 2.0s and 27m/s after 4.0s at what time will it come to a stop

1 answer:

lyudmila [28]3 years ago

8 0

Answer:

a1 = (v2 - v1) / t1 = (36 - 45) m/s / 2 s = -4.5 m/s^2

a2 = (v3 - v2) / t2 = (27 - 36) m/s / 2 s = -4.5 m / s^2

So the vehicle is apparently uniformly decelerating at -4.5 m/s^2

t = (vf - vi) / a = (0 - 45) m/s / -4.5 m/s^2 = 10 sec

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Plants absorb water from the soil through their roots. _______ of this water is used for photosynthesis; _______ of this water e

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THE CORRECT CHOICE IS B

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2 years ago

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An airplane flying at an altitude of 6 miles passes directly over a radar antenna. When the airplane is 10 miles away (s = 10),

Novosadov [1.4K]

Answer:

Explanation:

Given

altitude of the Plane $h=6\ miles$

When Airplane is $s=10\ miles$ away

Distance is changing at the rate of $\frac{\mathrm{d} s}{\mathrm{d} t}=290\ mph$

From diagram we can write as

$h^2+x^2=s^2$

differentiate above equation w.r.t time

$2h\frac{\mathrm{d} h}{\mathrm{d} t}+2x\frac{\mathrm{d} x}{\mathrm{d} t}=2s\frac{\mathrm{d} s}{\mathrm{d} t}$

as altitude is not changing therefore $\frac{\mathrm{d} h}{\mathrm{d} t}=0$

$0+x\frac{\mathrm{d} x}{\mathrm{d} t}=s\frac{\mathrm{d} s}{\mathrm{d} t}$

at $s=10\ miles\ and\ h=6\ miles$

substitute the value we get $x=\sqrt{10^2-6^2}=8\ miles$

$8\times \frac{\mathrm{d} x}{\mathrm{d} t}=10\times 290$

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3 years ago

The magnetic field in a solenoid is . A circular wire of radius 8 cm is concentric with a solenoid of radius 2 cm and length d =

defon

Answer:

6.03 mV

Explanation:

length of solenoid, L = 2 m, N = 12000, di/dt = 40 A/s,

Magnetic field due to solenoid

B = μ0 n i = μ0 N i / L

dB/dt = μ0 N / L x di / dt

dB /dt = (4 x 3.14 x 10^-7 x 12000 x 40) / 2 = 0.3 T/s

Induced emf, e = rate of change of magnetic flux

e = dΦ / dt = A x dB / dt

e = 3.14 x 0.08 x 0.08 x 0.3 = 6.03 x 10^-3 V = 6.03 mV

7 0

3 years ago

Find the ratio of the lengths of the two mathematical pendulums, if the ratio of periods is 1.5

juin [17]

Answer:

The ratio of lengths of the two mathematical pendulums is 9:4.

Explanation:

It is given that,

The ratio of periods of two pendulums is 1.5

Let the lengths be L₁ and L₂.

The time period of a simple pendulum is given by :

$T=2\pi \sqrt{\dfrac{l}{g}}$

or

$T^2=4\pi^2\dfrac{l}{g}\\\\l=\dfrac{T^2g}{4\pi^2}$

Where

l is length of the pendulum

$l\propto T^2$

or

$\dfrac{l_1}{l_2}=(\dfrac{T_1}{T_2})^2$ ....(1)

ATQ,

$\dfrac{T_1}{T_2}=1.5$

Put in equation (1)

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So, the ratio of lengths of the two mathematical pendulums is 9:4.

3 0

2 years ago

An Atwood machine consists of two masses, mA = 6.8 kg and mB = 8.0 kg , connected by a cord that passes over a pulley free to ro

Lisa [10]

To solve this problem it is necessary to apply the concewptos related to Torque, kinetic movement and Newton's second Law.

By definition Newton's second law is described as

F= ma

Where,

m= mass

a = Acceleration

Part A) According to the information (and as can be seen in the attached graph) a sum of forces is carried out in mass B, it is obtained that,

$\sum F = m_b a$

$m_Bg-T_B = m_Ba$

$T_B = m_Bg-m_Ba$

In the case of mass A,

$\sum F = m_A a$

$T_A = m_Ag-m_Aa$

Making summation of Torques in the Pulley we have to

$\sum\tau = I\alpha$

$T_BR_0-T_AR_0=I\alpha$

$T_B-T_A=I\frac{a}{R^2_0}$

Replacing the values previously found,

$(m_Bg-m_Ba )-(m_Ag-m_Aa )=I\frac{a}{R^2_0}$

$(m_B-m_A)g-(m_B+m_A)a=I\frac{a}{R^2_0}$

$a = \frac{(m_B-m_A)g}{\frac{I}{R_0^2}+(m_B+m_A)}$

$a = \frac{(m_B-m_A)g}{\frac{MR^2_0^2/2}{R_0^2}+(m_B+m_A)}$

$a =\frac{(m_B-m_A)g}{\frac{M}{2}+(m_B+m_A)}$

Replacing with our values

$a =\frac{(8-6.8)(9.8)}{\frac{0.8}{2}+(8+6.8)}$

$a=0.7736m/s^2$

PART B) Ignoring the moment of inertia the acceleration would be given by

$a' =\frac{(m_B-m_A)g}{(m_B+m_A)}$

$a' =\frac{(8-6.8)(9.8)}{(8+6.8)}$

$a' = 0.7945$

Therefore the error would be,

$\%error = \frac{a'-a}{a}*100$

$\%error = \frac{0.7945-0.7736}{0.7736}*100$

$\%error = 2.7%$

8 0

3 years ago