ANSWER
T₂ = 10.19N
EXPLANATION
Given:
• The mass of the ball, m = 1.8kg
First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,
In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,
The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,
We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,
Solve for T₁,
Now, we use the second equation to find the tension in the horizontal string,
Solve for T₂,
Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.
Answer:
A. The time taken for the car to stop is 3.14 secs
B. The initial velocity is 81.64 ft/s
Explanation:
Data obtained from the question include:
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Final velocity (V) = 0
Time (t) =?
Initial velocity (U) =?
A. Determination of the time taken for the car to stop.
Let us obtain an express for time (t)
Acceleration (a) = Velocity (V)/time(t)
a = V/t
Velocity (V) = distance (s) /time (t)
V = s/t
a = s/t^2
Cross multiply
a x t^2 = s
Divide both side by a
t^2 = s/a
Take the square root of both side
t = √(s/a)
Now we can obtain the time as follow
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Time (t) =..?
t = √(s/a)
t = √(256/26)
t = 3.14 secs
Therefore, the time taken for the car to stop is 3.14 secs
B. Determination of the initial speed of the car.
V = U + at
Final velocity (V) = 0
Deceleration (a) = –26ft/s2
Time (t) = 3.14 sec
Initial velocity (U) =.?
0 = U – 26x3.14
0 = U – 81.64
Collect like terms
U = 81.64 ft/s
Therefore, the initial velocity is 81.64 ft/s
Answer:
Bone
Explanation:
Diagnostic radiology include the use of non-invasive imaging scans to diagnose a patient.
The voltages used in diagnostic tubes range from roughly 20 kV to 150 kV and thus the highest energies of the X-ray photons range from roughly 20 keV to 150 keV.
The tests and equipment used sometimes involves low doses of radiation to create highly detailed images of an area.
a) PE=mgh=0.2*9.8*1.2=2.352 J
b) KE=PE=2.352 J
c) 
m/s
Answer:
v_g,i = 1.208 m/s
Explanation:
We are given;
Mass of girl; m_g = 47.2 kg
Mass of plank; m_p = 177 kg
Let the velocity of girl to ice be v_g,i
Let the velocity of plank to ice be v_p,i
Since the velocity of the girl is 1.53 m/s relative to the plank, then;
v_g,i + v_p,i = 1.53
From conservation of momentum;
m_g × v_g,i = m_p × v_p,i
Thus;
47.2(v_g,i) = 177(v_p,i)
Dividing both sides by 47.2 gives;
v_g,i = 3.75(v_p,i)
v_pi = (v_g,i)/3.75
Thus, from v_g,i + v_p,i = 1.53, we have;
v_g,i + ((v_g,i)/3.75) = 1.53
v_g,i(1 + 1/3.75) = 1.53
1.267v_g,i = 1.53
v_g,i = 1.53/1.267
v_g,i = 1.208 m/s
