Inserting the formulas you found for Xman(t) and Xbus(t) into the conditionXman(tcatch)=Xbus(tcatch) , you obtain the following-
1 answer:
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Answer:
P = 2439.5 W = 2.439 KW
Explanation:
First, we will find the mass of the water:
Mass = (Density)(Volume)
Mass = m = (1 kg/L)(10 L)
m = 10 kg
Now, we will find the energy required to heat the water between given temperature limits:
E = mCΔT
where,
E = energy = ?
C = specific heat capacity of water = 4182 J/kg.°C
ΔT = change in temperature = 95°C - 25°C = 70°C
Therefore,
E = (10 kg)(4182 J/kg.°C)(70°C)
E = 2.927 x 10⁶ J
Now, the power required will be:
where,
t = time = (20 min)(60 s/1 min) = 1200 s
Therefore,
<u>P = 2439.5 W = 2.439 KW</u>
<h2>Answer: 12 s</h2>
Explanation:
The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.
In this sense, the main movement equation in the Y axis is:
(1)
Where:
is the instrument's final position
is the instrument's initial position
is the instrument's initial velocity
is the time the parabolic movement lasts
is the acceleration due to gravity at the surface of planet X.
As we know and
when the object hits the ground, equation (1) is rewritten as:
(2)
Finding :
(3)
(4)
(5)
Finally:
Answer:
a) b) d)
Explanation:
The question is incomplete. The Complete question might be
In an inertial frame of reference, a series of experiments is conducted. In each experiment, two or three forces are applied to an object. The magnitudes of these forces are given. No other forces are acting on the object. In which cases may the object possibly remain at rest? The forces applied are as follows: Check all that apply.
a)2 N; 2 N
b) 200 N; 200 N
c) 200 N; 201 N
d) 2 N; 2 N; 4 N
e) 2 N; 2 N; 2 N
f) 2 N; 2 N; 3 N
g) 2 N; 2 N; 5 N
h ) 200 N; 200 N; 5 N
For th object to remain at rest, sum of all forces must be equal to zero. Use minus sign to show opposing forces
a) 2+(-2)=0 here minus sign is to show the opposing firection of force
b) 200+(-200)=0
c) 200+(-201)0
d) 2+2+(-4)=0
e) 2+2+(-2)0
f) 2+2+(-3) 0; 2+(-2)+3
0
g) 2+2+(-5)0; 2+(-2)+5
0
h)200 + 200 +(-5)0; 200+(-200)+5
0