Answer:
The ball land at 3.00 m.
Explanation:
Given that,
Speed = 40 m/s
Angle = 35°
Height h = 1 m
Height of fence h'= 12 m
We need to calculate the horizontal velocity
Using formula of horizontal velocity
[tex]V_{x}=V_{i}\cos\theta[/tex]
[tex]V_{x}=40\times\cos35[/tex]
[tex]V_{x}=32.76\ m/s[/tex]
We need to calculate the time
Using formula of time
[tex]t = \dfrac{d}{v}[/tex]
[tex]t=\dfrac{130}{32.76}[/tex]
[tex]t=3.96\ sec[/tex]
We need to calculate the vertical velocity
[tex]v_{y}=v_{y}\sin\theta[/tex]
[tex]v_{y}=40\times\sin35[/tex]
[tex]v_{y}=22.94\ m/s[/tex]
We need to calculate the vertical position
Using formula of distance
[tex]y(t)=y_{0}+V_{i}t+\dfrac{1}{2}gt^2[/tex]
Put the value into the formula
[tex]y(3.96)=1+22.94\times3.96+\dfrac{1}{2}\times(-9.8)\times(3.96)^2[/tex]
[tex]y(3.96)=15.00\ m[/tex]
We need to calculate the distance
[tex]s = y-h'[/tex]
[tex]s=15.00-12[/tex]
[tex]s=3.00\ m[/tex]
Hence, The ball land at 3.00 m.
A kangaroo can jump over an object 2.46 m high. (a) Calculate its vertical speed (in m/s) when it leaves the ground.
m/s
(b) How long (in s) is it in the air?
Answer:
a) 6.95 m/s
b) 1.42 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
[tex]v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 2.46-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 2.46}\\\Rightarrow u=6.95\ m/s[/tex]
a) The vertical speed when it leaves the ground. is 6.95 m/s
[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-6.95}{-9.81}\\\Rightarrow t=0.71\ s[/tex]
Time taken to reach the maximum height is 0.71 seconds
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 2.46=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.46\times 2}{9.81}}\\\Rightarrow t=0.71\ s[/tex]
Time taken to reach the ground from the maximum height is 0.71 seconds
b) Time it stayed in the air is 0.71+0.71 = 1.42 seconds
A 1.10 μF capacitor is connected in series with a 1.92 μF capacitor. The 1.10 μF capacitor carries a charge of +10.1 μC on one plate, which is at a potential of 51.5 V. (a) Find the potential on the negative plate of the 1.10 μF capacitor. (b) Find the equivalent capacitance of the two capacitors.
Answer:
(a). The potential on the negative plate is 42.32 V.
(b). The equivalent capacitance of the two capacitors is 0.69 μF.
Explanation:
Given that,
Charge = 10.1 μC
Capacitor C₁ = 1.10 μF
Capacitor C₂ = 1.92 μF
Capacitor C₃ = 1.10 μF
Potential V₁ = 51.5 V
Let V₁ and V₂ be the potentials on the two plates of the capacitor.
(a). We need to calculate the potential on the negative plate of the 1.10 μF capacitor
Using formula of potential difference
[tex]V_{1}=\dfrac{Q}{C_{1}}[/tex]
Put the value into the formula
[tex]V_{1}=\dfrac{10.1 \times10^{-6}}{1.10\times10^{-6}}[/tex]
[tex]V_{1}=9.18\ V[/tex]
The potential on the second plate
[tex]V_{2}=V-V_{1}[/tex]
[tex]V_{2}=51.5 -9.18[/tex]
[tex]V_{2}=42.32\ v[/tex]
(b). We need to calculate the equivalent capacitance of the two capacitors
Using formula of equivalent capacitance
[tex]C=\dfrac{C_{1}\timesC_{2}}{C_{1}+C_{2}}[/tex]
Put the value into the formula
[tex]C=\dfrac{1.10\times10^{-6}\times1.92\times10^{-6}}{(1.10+1.92)\times10^{-6}}[/tex]
[tex]C=6.99\times10^{-7}\ F[/tex]
[tex]C=0.69\ \mu F[/tex]
Hence, (a). The potential on the negative plate is 42.32 V.
(b). The equivalent capacitance of the two capacitors is 0.69 μF.
To practice Problem-Solving Strategy 21.1 Coulomb's Law. Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 2.3 cm . Two of the particles have a negative charge: q1 = -6.7 nC and q2 = -13.4 nC . The remaining particle has a positive charge, q3 = 8.0 nC . What is the net electric force acting on particle 3 due to particle 1 and particle 2?
Answer:
241.27 N
Explanation:
Both the negative charge will pull the positive charge towards it . Let the forces be F₁ and F₂
F₁ = [tex]\frac{9\times10^9\times8\times6.7\times10^{-18}}{(2.3\times10^{-2})^2}[/tex]
F₁ = 91.19 X 10⁻⁵ N
F₂=[tex]\frac{9\times10^9\times8\times13.4\times10^{-18}}{(2.3\times10^{-2})^2}[/tex]
F₂ = 182.38 X 10⁻⁵ N
F₁ and F₂ act at 60 degree so their resultant will be calculated as follows
R² = (91.19)² +(182.38)² + 2 X 91.19 X 182.38 Cos 60
R² = 58209.30
R = 241.27 N
A pipe carrying 20°C water has a diameter of 3.8 cm. Estimate the maximum flow speed (cm/s) if the flow must be streamline. (The density of water is 1.00 x 10^3 kg/m^3.)
Answer:
maximum flow speed is 5.278 cm/s
Explanation:
given data
temperature = 20°C
diameter d = 3.8 cm
density of water ρ = 1 × 10³ kg/m³
to find out
maximum flow speed
solution
we have given steam line flow so it is laminar flow
and for laminar flow we know reynolds number is Re= 2000
so
Re = [tex]\frac{\rho*v*d}{\mu}[/tex] ............1
here μ is dynamic viscosity = 0.001003 for 20°C for water
put all value in equation 1
Re = [tex]\frac{\rho*v*d}{\mu}[/tex]
2000 = [tex]\frac{1000*v*3.8*10^{-2}}{0.001003}[/tex]
solve it we get v
v = 0.0527 m/s
so maximum flow speed is 5.278 cm/s
Two bicyclist, originally separated by a distance of 20 miles, are each traveling at a uniform speed of 10 miles per hour toward each other down a long straight east-west aligned road. As the bikes travel, a bumble bee flying at a uniform speed of 25 miles per hour travels from the front wheel of one bike to the other, instantaneously reversing its course each time it encounters one of the bicycle wheels. Alas, eventually the two bicycles collide, crushing the poor bumble bee between their front wheels. What distance does the bee travel before his demise?
Answer:
D = 25 miles
Explanation:
To solve this problem, we just need to know how much time it took both bicyclists to collide and that will be the same amount of time that the bee flew at 25miles per hour. With those values we could calculate the distance it traveled.
Since both bicyclists collide, we know that Xa=Xb, so:
Xa = V*t = 10*t and Xb = 20 - V*t = 20 - 10*t
10*t = 20 - 10*t Solving for t:
t = 1 hour Now we can calculate the distance for the bee:
D = Vbee * t = 25 * 1 = 25 miles