Answer:
The new speed of the ball is 176.43 m/s
Explanation:
Given;
mass of the ball, m = 7 kg
initial speed of the ball, u = 5 m/s
applied force, F = 300 N
time of force action on the ball, t = 4 s
Apply Newton's second law of motion;
![F = ma = \frac{m(v-u)}{t}\\\\m(v-u) = Ft\\\\v-u = \frac{Ft}{m}\\\\v = \frac{Ft}{m} + u](https://tex.z-dn.net/?f=F%20%3D%20ma%20%3D%20%5Cfrac%7Bm%28v-u%29%7D%7Bt%7D%5C%5C%5C%5Cm%28v-u%29%20%3D%20Ft%5C%5C%5C%5Cv-u%20%3D%20%5Cfrac%7BFt%7D%7Bm%7D%5C%5C%5C%5Cv%20%3D%20%20%5Cfrac%7BFt%7D%7Bm%7D%20%2B%20u)
where;
v is new speed of the ball
![v = \frac{Ft}{m} + u\\\\v =\frac{300*4}{7} + 5\\\\v = 176.43 \ m/s](https://tex.z-dn.net/?f=v%20%3D%20%20%5Cfrac%7BFt%7D%7Bm%7D%20%2B%20u%5C%5C%5C%5Cv%20%3D%5Cfrac%7B300%2A4%7D%7B7%7D%20%2B%205%5C%5C%5C%5Cv%20%3D%20176.43%20%5C%20m%2Fs)
Therefore, the new speed of the ball is 176.43 m/s
Answer:
2.572 m/s²
Explanation:
Convert the given initial velocity and final velocity rates to m/s:
- 65 km/h → 18.0556 m/s
- 35 km/h → 9.72222 m/s
The motorboat's displacement is 45 m during this time.
We are trying to find the acceleration of the boat.
We have the variables v₀, v, a, and Δx. Find the constant acceleration equation that contains all four of these variables.
Substitute the known values into the equation.
- (9.72222)² = (18.0556)² + 2a(45)
- 94.52156173 = 326.0046914 + 90a
- -231.4831296 = 90a
- a = -2.572
The magnitude of the boat's acceleration is |-2.572| = 2.572 m/s².
Answer:
Explanation:
Expression for time period of pendulum is given as follows
![T=2\pi\sqrt{\frac{l}{g} }](https://tex.z-dn.net/?f=T%3D2%5Cpi%5Csqrt%7B%5Cfrac%7Bl%7D%7Bg%7D%20%7D)
where l is length of pendulum and g is acceleration due to gravity .
Putting the given values for first place
![2.67=2\pi\sqrt{\frac{l}{9.77} }](https://tex.z-dn.net/?f=2.67%3D2%5Cpi%5Csqrt%7B%5Cfrac%7Bl%7D%7B9.77%7D%20%7D)
Putting the values for second place
![T=2\pi\sqrt{\frac{l}{9.81} }](https://tex.z-dn.net/?f=T%3D2%5Cpi%5Csqrt%7B%5Cfrac%7Bl%7D%7B9.81%7D%20%7D)
Dividing these two equation
![\frac{T}{2.67} =\sqrt{\frac{9.77}{9.81} }](https://tex.z-dn.net/?f=%5Cfrac%7BT%7D%7B2.67%7D%20%3D%5Csqrt%7B%5Cfrac%7B9.77%7D%7B9.81%7D%20%7D)
T = 2.66455 s.
Answer:
Number of electrons, ![n=5.56\times 10^{10}](https://tex.z-dn.net/?f=n%3D5.56%5Ctimes%2010%5E%7B10%7D)
Explanation:
Given that,
Charge on the fur, ![q=8.9\ nC=8.9\times 10^{-9}\ C](https://tex.z-dn.net/?f=q%3D8.9%5C%20nC%3D8.9%5Ctimes%2010%5E%7B-9%7D%5C%20C)
A piece of amber is charged by rubbing with a piece of fur. We need to find the electrons were added to the amber. It can be calculated using the quantization of charge as
q = n × e
Where
n is the number of electrons
e is the charge on electron
![n=5.56\times 10^{10}\ electrons](https://tex.z-dn.net/?f=n%3D5.56%5Ctimes%2010%5E%7B10%7D%5C%20electrons)
So,
number of electrons are added to the amber. Therefore, this is the required solution.
<span>The Earth’s orbit is a nearly circular ellipse.
</span><span>The Sun is located at one of the two focal points.</span>
The Earth moves around the Sun in an orbit that is almost but not quite circular. As Kepler proved in the seventeenth century, the orbit is actually an ellipse. A parameter called the eccentricity (e) defines the degree of departure from a circle. A value of e=0 would indicate a circle whereas a value of e=0.9 would indicate a very elongated ellipse. The eccentricity of the Earth's orbit is currently e=0.0167.