Answer:
185.25 m/s
Explanation:
consider the motion of the combination of bullet and block after the collision
v₀ = initial speed just after the collision
v' = final speed = 0 m/s
μ = Coefficient of friction = 0.6
g = acceleration due to gravity = 9.8 m/s²
a = acceleration of the combination = - μ g = - (0.6) (9.8) = - 5.88 m/s²
d = stopping distance = 13 m
using the kinematics equation
v'² = v₀² + 2 a d
0² = v₀² + 2 (- 5.88) (13)
v₀ = 12.4 m/s
m = mass of the bullet = 9.9 g = 0.0099 kg
M = mass of the wood = 138 g = 0.138 kg
v = speed of bullet before collision
v₀ = speed of combination after the collision = 12.4 m/s
Using conservation of momentum
m v = (m + M) v₀
(0.0099) v = (0.0099 + 0.138) (12.4)
v = 185.25 m/s
Answer:
7] Force = mass × acceleration
Force = 2 × 5
<u>Force = 10 N</u>
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8] Velocity = acceleration due to gravity × time taken
Velocity = 9.8 × 12
<u>Velocity = 117.6 m/s</u>
(a) Determine the circumference of the Earth through the equation,
C = 2πr
Substituting the known values,
C = 2π(1.50 x 10¹¹ m)
C = 9.424 x 10¹¹ m
Then, divide the answer by time which is given to a year which is equal to 31536000 s.
orbital speed = (9.424 x 10¹¹ m)/31536000 s
orbital speed = 29883.307 m/s
Hence, the orbital speed of the Earth is ~29883.307 m/s.
(b) The mass of the sun is ~1.9891 x 10³⁰ kg.
I believe the correct answer from the choices listed above is option 4. Base from the passage given above, it is very clear that the Campylobacter is a bacteria used to test in <span>identifying drug-resistant bacteria. Hope this answers the question. Have a nice day.</span>
Answer:
Answered
Explanation:
A) The work done by gravity is zero because displacement and the gravitational force are perpendicular to each other.
W= FS cosθ
θ= 90 ⇒cos90 = 0 ⇒W= 0
B) work done by tension
W= Tcosθ×S= 5cos30×2.30= 10J
C) Work done by friction force
W= f×s=1×2.30= 2.30 J
D) Work done by normal force is Zero because the displacement and the normal force are perpendicular to each other.
E) The net work done= Work done by tension in the rope - frictional work
=10-2.30= 7.7 J