A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C.
1 answer:
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<h2>Answer:</h2>
(c) 5m/s²
<h2>Explanation:</h2>Total acceleration (a) of a particle in a circular motion is the vector sum of the radial or centripetal acceleration () of the particle and the tangential acceleration (
) of the particle and its magnitude can be calculated as follows;
a = ---------------------(i)
<em>But;</em>
=
------------------------------(ii)
Where;
v = instantaneous velocity
r = radius of the circular path of motion
<em>From the question;</em>
v = 30m/s
r = 300m
(i) First let's calculate the centripetal acceleration by substituting the values above into equation (ii) as follows;
=
=
= 3m/s²
(ii) From the question, the velocity of the particle is increasing at a constant rate of 4m/s² and that is the tangential acceleration , of the particle. i.e;
= 4m/s²
(iii) Now substitute the values of and
into equation (i) as follows;
a =
a =
a =
a = 5m/s²
Therefore, the magnitude of its total acceleration a, is 5m/s²
Answer:
Charge,
Explanation:
It is given that,
Electric field strength, E = 180000 N/C
Distance from a small object, r = 2.8 cm = 0.028 m
Electric field at a point is given by :
Q is the charge on an object
So, the charge on the object is . Hence, this is the required solution.
Answer:
r = 4.21 10⁷ m
Explanation:
Kepler's third law It is an application of Newton's second law where the forces of the gravitational force, obtaining
T² = ( ) r³ (1)
in this case the period of the season is
T₁ = 93 min (60 s / 1 min) = 5580 s
r₁ = 410 + 6370 = 6780 km
r₁ = 6.780 10⁶ m
for the satellite
T₂ = 24 h (3600 s / 1h) = 86 400 s
if we substitute in equation 1
T² = K r³
K = T₁²/r₁³
K =
K = 9.99 10⁻¹⁴ s² / m³
we can replace the satellite values
r³ = T² / K
r³ = 86400² / 9.99 10⁻¹⁴
r = ∛(7.4724 10²²)
r = 4.21 10⁷ m
this distance is from the center of the earth