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Afina-wow [57]
1 year ago
8

A plane flying at 70.0 m/s suddenly stalls. If the acceleration during the stall is 9.8 m/s2 directly downward, the stall lasts

5.0 s, and the plane was originally climbing at 25° to the horizontal, what is the velocity after the stall?
Physics
1 answer:
tino4ka555 [31]1 year ago
4 0

Answer:

v = 66.4 m/s

Explanation:

As we know that plane is moving initially at speed of

v = 70 m/s

now we have

v_x = 70 cos25

v_x = 63.44 m/s

v_y = 70 sin25

v_y = 29.6 m/s

now in Y direction we can use kinematics

v_y = v_i + at

v_y = 29.6 - (9.81 \times 5)

v_y = -19.5 m/s

since there is no acceleration in x direction so here in x direction velocity remains the same

so we will have

v = \sqrt{v_x^2 + v_y^2}

v = \sqrt{63.44^2 + 19.5^2}

v = 66.4 m/s

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A & B

Observe the path of the light ray as it passes through the lenses as shown in the attached images. Concave lenses diverge light rays while the convex lens converges the light rays.

Explanation:

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6 0
2 years ago
Read 2 more answers
A car is pulled with a force of 10,000 N. The car's mass is 1267 kg. But, the car covers 394.6 m in 15 seconds
solong [7]

A. Formula: F=ma or F/m=a

10,000N/1,267kg≈7.9m/s^{2}

B. Formula: a=\frac{V-V_{0} }{t} and s=d/t

speed= 394.6/15

s=26.3m/s

a=\frac{26.3-0}{15}

a=1.75m/s^{2}

C. 7.9-1.75=difference of 6.15m/s^{2}

D. The force that most likely caused this difference is friction forces

3 0
1 year ago
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1. A2 .7-kg copper block is given an initial speed of 4.0m/s on a rough horizontal surface. Because of friction, the block final
tatyana61 [14]

Answer:

A. Increase in temperature is 0.0176 degree Celsius. b. the remaining energy will be lost.

Explanation:

The mass of copper block = 7kg

Initial speed = 4.0 m/s

Specific heat of copper = 0.385 j/g degree Celcius.

a. The increase in temperature is calculated below:

\text{Change in kinetic energy} = \frac{1}{2}mv^{2} \\=  \frac{1}{2} \times 2.7 \times 4^{2} = 21.6 J \\

85% of energy is converted into internal energy.

mc\DeltaT = 21.6 \times 0.85 \\2.7 \times 385 \times \Delta T = 21.6 \times 0.85 \\\Delta T = 0.0176 degree \ celsius

b. The remaining  15 per cent of kinetic energy will be lost and it will be changed into other forms.

7 0
1 year ago
An LR circuit contains an ideal 60-V battery, a 42-H inductor having no resistance, a 24-ΩΩ resistor, and a switch S, all in ser
s2008m [1.1K]

Answer:

1.6 s

Explanation:

To find the time in which the potential difference of the inductor reaches 24V you use the following formula:

V_L=V_oe^{-\frac{Rt}{L}}

V_o: initial voltage = 60V

R: resistance = 24-Ω

L: inductance = 42H

V_L: final voltage = 24 V

You first use properties of the logarithms to get time t, next, replace the values of the parameter:

\frac{V_L}{V_o}=e^{-\frac{Rt}{L}}\\\\ln(\frac{V_L}{V_o})=-\frac{Rt}{L}\\\\t=-\frac{L}{R}ln(\frac{V_L}{V_o})\\\\t=-\frac{42H}{24\Omega}ln(\frac{24V}{60V})=1.6s

hence, after 1.6s the inductor will have a potential difference of 24V

3 0
2 years ago
8.4-1 Consider a magnetic field probe consisting of a flat circular loop of wire with radius 10 cm. The probe’s terminals corres
Vlad1618 [11]

Answer:

B_o = 1.013μT

Explanation:

To find B_o you take into account the formula for the emf:

\epsilon=-\frac{d\Phi_b}{dt}=-\frac{dBAcos\theta}{dt}=-Acos\theta\frac{dB}{dt}

where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.

By applying the derivative you obtain:

\epsilon=-Acos\theta (2\pi f) B_ocos(2\pi f t+ \alpha)

when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:

\epsilon=2\pi fAB_o=2\pi (100*10^3Hz)(\pi (0.1m)^2)B_o=19739.20Hzm^2B_o\\\\B_o=\frac{20*10^{-3}V}{19739.20Hzm^2}=1.013*10^{-6}T=1.013\mu T

hence, B_o = 1.013μT

6 0
2 years ago
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