A car turns from a road into a parking lot and into an available parking space. The car’s initial velocity is 4 m/s [E 45° N]. The car’s velocity just before the driver decreases speed is 4 m/s [E 10° N]. The turn takes 3s. What's the average acceleration of the car during the turn? The answer should have directions with an angle.
Answers
Write the velocity vectors in component form.
• initial velocity:
v₁ = 4 m/s at 45º N of E
v₁ = (4 m/s) (cos(45º) i + sin(45º) j)
v₁ ≈ (2.83 m/s) i + (2.83 m/s) j
• final velocity:
v₂ = 4 m/s at 10º N of E
v₂ = (4 m/s) (cos(10º) i + sin(10º) j)
v₂ ≈ (3.94 m/s) i + (0.695 m/s) j
The average acceleration over this 3-second interval is then
a = (v₂ - v₁) / (3 s)
a ≈ (0.370 m/s²) + (-0.711 m/s²)
with magnitude
||a|| = √[(0.370 m/s²)² + (-0.711 m/s²)²] ≈ 0.802 m/s²
and direction θ such that
tan(θ) = (-0.711 m/s²) / (0.370 m/s²) ≈ -1.92
→ θ ≈ -62.5º
which corresponds to an angle of about 62.5º S of E, or 27.5º E of S. To use the notation in the question, you could say it's E 62.5º S or S 27.5º E.
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Answers
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Thank you and please rate me as brainliest as it will help me to level up
A. 2000 J
B. 75,000 J
C. 120,000 J
D. 300,000 J
Answers
The electrical energy used by a 400 W toaster that is operating for 5 minutes will be 120,000 J.Option C is correct.
What is the power output?
The rate of the work done is called the power output. It is denoted by P.Its unit of a watt. It is the ratio of the work done or the enrgy to the time period.
The given data in the problem is;
E is the electrical energy
P is the power output = 400 W
t is the time period = 5 minutes
The power output is given as;
Hence the electrical energy used by a 400 W toaster that is operating for 5 minutes will be 120,000 J.Option C is correct.
To learn more about the power output refer to the link;
Answer:
The answer is C. 120,000 J.
Explanation:
These radio waves travel at a speed of 3.00 x 108 m/s.
What is the wavelength of these radio waves?
Answers
Wavelength = 3.00x108/9.05x107=
3.3x10 risen to the power of -1
Answers
Answer:
10.284 kgm/s
4.674545 kN
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
m = Mass of baselball
Impulse
Impulse of the baseball is 10.284 kgm/s
Impulse is also given by
The magnitude of the average force exerted by the bat on the ball is 4.674545 kN
Answers
Answer:
(a) 4.875 m
(b) 5.72 rad/s
(c) 0.858 m/s
Explanation:
(a) Assuming constant angular speed, the angular distance the ball would have traveled after 5s at the rate of 6.5 rad/s is
6.5 * 5 = 32.5 rad
With radius of 0.15m, the linear distance it would have traveled is
32.5 * 0.15 = 4.875 m
(b)The angular velocity of the ball after 0.65s when subjected to an angular acceleration of -1.2 rad/s is
6.5 - 1.2*0.65 = 5.72 rad/s
(c)The linear speed of the ball is the product of the angular speed and radius 0.15 m
5.72 * 0.15 = 0.858 m/s
Answers
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Explanation:
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