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Complete Question:
a. A hawk flies in a horizontal arc of radius 11.3 m at a constant speed of 5.7 m/s. Find its centripetal acceleration.
Answer in units of m/s2
b. It continues to fly along the same horizontal arc but increases its speed at the rate of 1.34 m/s2. Find the magnitude of acceleration under
these new conditions.
Answer in units of m/s2
Answer:
a. 2.875m/s²
b. 3.172m/s²
Explanation:
a. The formula for centripetal acceleration = (speed²) ÷ radius
Centripetal acceleration = (5.7m/s)²÷ 11.3m
Centripetal acceleration = 2.875m/s²
b. Magnitude of acceleration can be calculated by finding the sum of the vectors for the both the centripetal acceleration and the increase in the speed rate.
Centripetal acceleration ( acceleration x) = 2.875m/s²
Increase in the speed rate ( acceleration n) = 1.34m/s²
Magnitude of acceleration = √a²ₓ + a²ₙ
=√( 2.875m/s²)²+ (1.34m/s²)²
= √ 10.06m/s²
= 3.172m/s²
According to Edge, the answers is
<h3>3100 V</h3>
and
<h3>200v</h3>
Velocity =displacement
Change in time
D=72km/hr
Time=20s
But the S.I unit of velocity is m/s so you woul have to change 72km/hr to m/s
Changing 72km to m
1 kilometer=1000meters
Then, 72 kilometers =?
72•1000/1
=72000m
Changing 72hours to seconds
If 1 hour = 3600 seconds
Then 72 hours=?
72•3600/1
=259200 seconds
Velocity =displacement
Change in time
V= 72,000
259,2005
=0.028m/s
Answer:
P = 1 x 10⁸ Pa
Explanation:
given,
radius = 2.0 ×10⁻¹⁰ m
Temperature
T = 300 K
Volume of gas molecule =
V = 33.51 x 10⁻³⁰ m³
we know,
P V = 1 . k T
k = 1.38 x 10⁻²³ J/K
P(33.51 x 10⁻³⁰) = 1 . (1.38 x 10⁻²³) x 300
P = 1.235 x 10⁸ Pa
for 1 significant figure
P = 1 x 10⁸ Pa
