A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110.m. Determine the acceleration
1 answer:
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Answer:
The acceleration of the both masses is 0.0244 m/s².
Explanation:
Given that,
Mass of one block = 602.0 g
Mass of other block = 717.0 g
Radius = 1.70 cm
Height = 60.6 cm
Time = 7.00 s
Suppose we find the magnitude of the acceleration of the 602.0-g block
We need to calculate the acceleration
Using equation of motion
Where, s = distance
t = time
a = acceleration
Put the value into the formula
Hence, The acceleration of the both masses is 0.0244 m/s².
Answer:
v = 5.9 x 10⁷ m/s
Explanation:
The kinetic energy of the electron in terms of potential difference is given as:
--------------- equation (1)
where,
e = charge on electron = 1.6 x 10⁻¹⁹ C
V = Potential Difference = 9.9 KV = 9900 Volts
The kinetic energy in general is given as:
--------- equation (2)
where,
m = mass of electron = 9.1 x 10⁻³¹ kg
v = speed of electron = ?
Therefore, comparing equation (1) and equation (2), we get:
<u>v = 5.9 x 10⁷ m/s</u>
Answer:
a) The angular acceleration of the beam is 0.5 rad/s²CW (direction clockwise due the tangential acceleration is positive)
b) The acceleration of point A is 3.25 m/s²
The acceleration of point E is 0.75 m/s²
Explanation:
a) The relative acceleration of B with respect to D is equal:
Where
aB = absolute acceleration of point B = 2.5 j (m/s²)
aD = absolute acceleration of point D = 1.5 j (m/s²)
(aB/D)n = relative acceleration of point B respect to D (normal direction BD) = 0, no angular velocity of the beam
(aB/D)t = relative acceleration of point B respect to D (tangential direction BD)
We have that
(aB/D)t = BDα
Where α = acceleration of the beam
BDα = 1 m/s²
Where
BD = 2
b) The acceleration of point A is:
(aA/D)t = ADαj
The acceleration of point E is:
(aE/D)t = -EDαj