A firefighting crew uses a water cannon that shoots water at 27.0 m/s at a fixed angle of 50.0 ∘ above the horizontal. The firefighters want to direct the water at a blaze that is 12.0 m above ground level. How far from the building should they position their cannon? There are two possibilities (d1Part A:
d1=_____m
Part B:
d2=______m
Answers
Answer:
Explanation:
In projectile motion , range of projectile is given by the expressions
R = u²sin2θ / g
where u is velocity of projectile.
u = 27 m/s θ = 50
12 = 27² sin 2θ / 9.8
sin 2θ = .16
θ = 9.2 / 2
= 4.6
When we place 90- θ in place of θ , in the formula of range , we get the same value of projectile. hence at 85.4 ° , the range will be same.
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A 133 kg horizontal platform is a uniform disk of radius 1.95 m and can rotate about the vertical axis through its center. A 62.7 kg person stands on the platform at a distance of 1.19 m from the center, and a 28.5 kg dog sits on the platform near the person 1.45 m from the center. Find the moment of inertia of this system, consisting of the platform and its population, with respect to the axis.
Answers
Answer:
-0.00152 V
Explanation:
Parameters given:
Diameter of the loop = 11 cm = 0.11m
Rate of change of magnetic field, dB/dt = 0.16 T/s
Radius of the loop = 0.055m
The area of the loop will be:
A = pi * r²
A = 3.142 * 0.055²
A = 0.0095 m²
The EMF induced in a loop of wire due to the presence of a changing magnetic field, dB, in a time interval, dt, is given as:
EMF = - N * A * dB/dt
In this case, there's only one loop, so N = 1.
Therefore:
EMF = -1 * 0.0095 * 0.16
EMF = -0.00152 V
The negative sign indicates that the current flowing through the loop acts opposite to the change in the magnetic field.
Answer:
The induced emf is 0.00152 V
Explanation:
Given data:
d = 11 cm = 0.11 m
The area is:
The induced emf is:
The negative indicates the direction of E.
Answers
Answer:
The Answer is 0.019998c
Explanation:
Please see the attached Picture for the answer.
Answers
Answer:TL;DR: 3.535 cm
Explanation:
Xcm = ΣxMoments/ΣMasses = (10*0 + 10*5)/(10+10) = 50/20 = 2.5 cm
by symmetry,
Ycm = 2.5 cm
The distance D from the point Xcm,Ycm to the origin is D = √(2.5²+2.5²) = 3.535 cm
Final answer:
The center of mass of the bent wire is approximately 11.18 cm from the bend.
Explanation:
In order to find the center of mass of the bent wire, we need to divide it into two segments: the horizontal segment and the vertical segment. The length of each segment is half of the total length of the wire, which is 20 cm, so each segment is 10 cm long.
The center of mass of the horizontal segment is located exactly at its middle point, which is 5 cm from the corner. The center of mass of the vertical segment is also located at its middle point, which is 10 cm from the corner. Since the horizontal and vertical segments are orthogonal, the distance from the bend to the center of mass of the bent wire is the hypotenuse of a right triangle with legs of length 5 cm and 10 cm. Using the Pythagorean theorem, we can calculate the distance:
d = sqrt(5^2 + 10^2) = sqrt(25 + 100) = sqrt(125) = 11.18 cm
Therefore, the center of mass of the bent wire is approximately 11.18 cm from the bend.
Answers
Answer:
d = 3.44 x 10⁸ m
Explanation:
The minimum distance required will be the distance from the centre of the earth to a point where gravitational intensity due to both earth and moon becomes equal . Once this point is reached , moon will attract the baseball on its own .
Let this distance be d from the centre of the earth
So GM / d² = G m / ( 3.82 x 10⁸ - d )²
M is mass of the earth , m is mass of the moon
M / m = ( d / 3.82 x 10⁸ - d )²
5.972 x 10²⁴ / 7.34 x 10²² = ( d / 3.82 x 10⁸ - d )²
81.36 = ( d / 3.82 x 10⁸ - d )²
9.02 = d / 3.82 x 10⁸ - d
34.45 x 10⁸ - 9.02 d = d
34.45 x 10⁸ = 10.02 d
d = 3.44 x 10⁸ m
Answers
Answer:
r₁/r₂ = 1/2 = 0.5
Explanation:
The resistance of a wire is given by the following formula:
R = ρL/A
where,
R = Resistance of wire
ρ = resistivity of the material of wire
L = Length of wire
A = Cross-sectional area of wire = πr²
r = radius of wire
Therefore,
R = ρL/πr²
FOR WIRE A:
R₁ = ρ₁L₁/πr₁² -------- equation 1
FOR WIRE B:
R₂ = ρ₂L₂/πr₂² -------- equation 2
It is given that resistance of wire A is four times greater than the resistance of wire B.
R₁ = 4 R₂
using values from equation 1 and equation 2:
ρ₁L₁/πr₁² = 4ρ₂L₂/πr₂²
since, the material and length of both wires are same.
ρ₁ = ρ₂ = ρ
L₁ = L₂ = L
Therefore,
ρL/πr₁² = 4ρL/πr₂²
1/r₁² = 4/r₂²
r₁²/r₂² = 1/4
taking square root on both sides:
r₁/r₂ = 1/2 = 0.5
Final answer:
The ratio of the radius of wire A to the radius of wire B is 1/2.
Explanation:
The resistance of a wire is given by the formula R = ρl/A, where R is resistance, ρ is resistivity, l is length, and A is the cross-sectional area of the wire. When the wire has a circular cross-section, the area can be calculated by the formula A = πr². The resistance of the wire then becomes: R = ρl/(πr²). If the resistance of wire A is four times that of wire B, we can set up the equation 4RB = RA. Substituting the expression for resistance, we get 4(ρl/(πrB²)) = ρl/(πrA²). Simplifying, we find that the ratio of the radius of wire A to the radius of wire B is one-half, or rA/rB = 1/2.
Learn more about Resistance and Radius Ratio here:
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Answers
Answer:
(a) t=3.87 s :time at which Kathy overtakes Stan
(b) d=37.36 m
(c) vf₁ = 15.097 m/s : Stan's final speed
vf₂ = 19.31 m/s : Kathy's final speed
Explanation:
kinematic analysis
Because Kathy and Stan move with uniformly accelerated movement we apply the following formulas:
vf= v₀+at Formula (1)
vf²=v₀²+2*a*d Formula (2)
d= v₀t+ (1/2)*a*t² Formula (3)
Where:
d:displacement in meters (m)
t : time in seconds (s)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Nomenclature
d₁: Stan displacement
t₁ : Stan time
v₀₁: Stan initial speed
vf₁: Stan final speed
a₁: Stan acceleration
d₂: car displacement
t₂ : Kathy time
v₀₂: Kathy initial speed
vf₂: Kathy final speed
a₂: Kathy acceleration
Data
v₀₁ = 0
v₀₂ = 0
a₁ = 3.1 m/s²
a₂= 4.99 m/s²
t₁ = (t₂ +1) s
Problem development
By the time Kathy overtakes Stan, the two will have traveled the same distance:
d₁ = d₂
t₁ = (t₂ +1)
We aplpy the Formula (3)
d₁ = v₀₁t₁ + (1/2)*a₁*t₁²
d₁ = 0 + (1/2)*(3.1)*t₁²
d₁ = 1.55*t₁² ; Stan's cinematic equation 1
d₂ = v₀₂t₂ + (1/2)*a₂*t₂²
d₂ = 0 + (1/2)*(4.99)*t₂²
d₂ = 2.495* t₂² : Kathy's cinematic equation 2
d₁ = d₂
equation 1=equation 2
1.55*t₁² = 2.495* t₂² , We replace t₁ = (t₂ +1)
1.55* (t₂ +1) ² =2.495* t₂²
1.55* (t₂² +2t₂+1) =2.495* t₂²
1.55*t₂²+1.55*2t₂+1.55 = 2.495* t₂²
1.55t₂²+3.1t₂+1.55=2.495t₂²
(2.495-1.55)t₂² - 3.1t₂ - 1.55 = 0
0.905t₂² - 3.1t₂ - 1.55 = 0 Quadratic equation
Solving the quadratic equation we have:
(a) t₂ = 3.87 s : time at which Kathy overtakes Stan
(b) Distance in which Kathy catches Stan
we replace t₂ = 3.87 s in equation 2
d₂ = 2.495*( 3.87)²
d₂ = 37.36 m
(c) Speeds of both cars at the instant Kathy overtakes Stan
We apply the Formula (1)
vf₁= v₀₁+a₁t₁ t₁ =( t₂+1 ) s=( 3.87 + 1 ) s = 4.87 s
vf₁= 0+3.1* 4.87
vf₁ = 15.097 m/s : Stan's final speed
vf₂ = v₀₂+a₂ t₂
vf₂ =0+4.99* 3.87
vf₂ = 19.31m/s : Kathy's final speed