Answer:
9) This is a case of deceleration
10)-0.8 ms-2
b) acceleration is the change in velocity with time
11)
a) 100 ms-1
b) 100 seconds
12) 10ms-1
13) more information is needed to answer the question
14) - 0.4 ms^-2
15) 0.8 ms^-2
Explanation:
The deceleration is;
v-u/t
v= final velocity
u= initial velocity
t= time taken
20-60/50 =- 40/50= -0.8 ms-2
11)
Since it starts from rest, u=0 hence
v= u + at
v= 10 ×10
v= 100 ms-1
b)
v= u + at but u=0
1000 = 10 t
t= 1000/10
t= 100 seconds
12) since the sprinter must have started from rest, u= 0
v= u + at
v= 5 × 2
v= 10ms-1
14)
v- u/t
10 - 20/ 25
10/25
=- 0.4 ms^-2
15)
a=v-u/t
From rest, u=0
8 - 0/10
a= 8/10
a= 0.8 ms^-2
We are given the mass of an <span>aluminum sculpture which is 145 kg and a horizontal force equal to 668 Newtons. The coefficient of friction can be determined by dividing the horizontal force by the weight of the object. In this case, 668 N / 145 * 9.8 equal to coeff of friction of 0.47</span>
Answer:
6.67×10¯⁹ A
Explanation:
From the question given above, the following data were obtained:
Quantity of electricity (Q) = 2 μC
Time (t) = 5 mins
Current (I) =?
Next, we shall convert 2 μC to C. This can be obtained as follow:
1 μC = 1×10¯⁶ C
Therefore,
2 μC = 2 μC × 1×10¯⁶ C / 1 μC
2 μC = 2×10¯⁶ C
Next, we shall convert 5 mins to seconds. This can be obtained as follow:
1 min = 60 secs
Therefore,
5 min = 5 min × 60 sec / 1 min
5 mins = 300 s
Finally, we shall determine the current in the circuit. This can be obtained as follow:
Quantity of electricity (Q) = 2×10¯⁶ C
Time (t) = 300 s
Current (I) =?
Q = It
2×10¯⁶ = I × 300
Divide both side by 300
I = 2×10¯⁶ / 300
I = 6.67×10¯⁹ A
Thus, the current in the circuit is 6.67×10¯⁹ A
Answer:
3300J
Explanation:
Work done is the energy that is lost by the skater
Formula for workdone = 1/2*mV^2
m = 66kg
V = 10m/s
Work done = 1/2 * 66 * 10^2
= 3300J
Answer:
1.19 m/s²
Explanation:
The frequency of the wave generated in the string in the first experiment is f = n/2l√T/μ were T = tension in string = mg were m = 1.30 kg weight = 1300 g , μ = mass per unit length of string = 1.01 g/m. l = length of string to pulley = l₀/2 were l₀ = lent of string. Since f is the second harmonic, n = 2, so
f = 2/2(l₀/2)√mg/μ = 2(√mg/μ)/l₀ (1)
Also, for the second experiment, the period of the wave in the string is T = 2π√l₀/g. From (1) l₀ = 2(√mg/μ)/f and from (2) l₀ = T²g/4π²
Equating (1) and (2) we ave
2(√mg/μ)/f = T²g/4π²
Making g subject of the formula
g = 2π√(2√(m/μ)/f)/T
The period T = 316 s/100 = 3.16 s
Substituting the other values into , we have
g = 2π√(2√(1300 g/1.01 g/m)/200 Hz)/3.16
g = 2π√(2 × 35.877/200 Hz)/3.16
g = 2π√(71.753/200 Hz)/3.16
g = 2π√(0.358)/3.16
g = 2π × 0.599/3.16
g = 1.19 m/s²
