Answers
Answer:
The highest of its trajectory = 0.45 m
Option C is the correct answer.
Explanation:
Considering vertical motion of cat:-
Initial velocity, u = 3.44 sin60 = 2.98 m/s
Acceleration , a = -9.81 m/s²
Final velocity, v = 0 m/s
We have equation of motion v² = u² + 2as
Substituting
v² = u² + 2as
0² = 2.98² + 2 x -9.81 x s
s = 0.45 m
The highest of its trajectory = 0.45 m
Option C is the correct answer.
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The spring is used in a spring gun to project a 10.0 g plastic ball. In the gun, the spring is compressed by 10.0 cm before the gun is fired. How fast is the ball traveling after the gun is fired?
A hovering mosquito is hit by a raindrop that is 50 times as massive and falling at 8.5 m/sm/s , a typical raindrop speed. How fast is the raindrop, with the attached mosquito, falling immediately afterward if the collision is perfectly inelastic
Answers
Answer:
θ=π/2
Explanation:
The definition of work is W = → F ⋅ → d = q E c o s θ d W=F→⋅d→=qEcosθd. So if no work is done, the displacement must be in the direction perpendicular to the force ie c o s θ = 0 → θ = π / 2 cosθ=0→θ=π/2
Final answer:
A charged particle can be displaced without any external work done on it in a uniform electric field when its movement is perpendicular to the direction of the electric field.
Explanation:
In a uniform electric field, the electric force is the same in every direction. Therefore, if a charge were to be displaced perpendicular to the original direction of the electric field (i.e., in the y or z direction), it would not encounter any extra electric forces. This means there would be no external work being done on the charge. When a charge is moved perpendicular to an electric field, the field does not affect it, and hence, no work is done by the field.
In other words, a charge can be displaced in this field without any external work being done on it when it is moved in a direction perpendicular to the uniform electric field, either in y-axis or z-axis, assuming the electric field is constant in the x-axis direction.
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Answers
To solve this problem it is necessary to apply the concepts related to Newton's second law and the equations of motion description for acceleration.
From the perspective of acceleration we have to describe it as
Where,
= Velocity
= time
At the same time by the Newton's second law we have that
F = ma
Where,
m = mass
a = Acceleration
Replacing the value of acceleration we have
Our values are given as,
Replacing we have,
Therefore the magnitude of the average force exerted on the ball by the club is 744.11N
Answers
The anchoring force needed to hold the elbow in place is 839.5 N.
How to find anchoring force?
To determine the anchoring force needed to hold the elbow in place, use the following steps:
Apply the conservation of momentum equation to the elbow:
∑F = ˙m(V₂ - V₁)
where:
˙m = mass flow rate
V₁ and V₂ = velocities at the inlet and outlet of the elbow, respectively
F = anchoring force
The mass flow rate is given by:
˙m = ρAV
where:
ρ = density of water (1000 kg/m³)
A = cross-sectional area of the elbow
V = velocity
The velocities at the inlet and outlet of the elbow can be calculated using the following equations:
V₁ = A₁V₁
V₂ = A₂V₂
where:
A₁ and A₂ = cross-sectional areas at the inlet and outlet of the elbow, respectively
Calculate the momentum flux correction factor at the inlet and outlet of the elbow:
β₁ = 1.03
β₂ = 1.03
Substitute all of the above equations into the conservation of momentum equation:
F = ˙m(V₂ - V₁)
F = ρA₁V₁²β₁ - ρA₂V₂²β₂
Calculate the velocity at the inlet of the elbow:
V₁ = A₂V₂/A₁
V₁ = (25 cm²/150 cm²)(V₂)
V₁ = 1/6 V₂
Substitute the above equation into the conservation of momentum equation:
F = ρA₁V₁²β₁ - ρA₂V₂²β₂
F = ρA₁[(1/6 V₂)²](1.03) - ρA₂V₂²(1.03)
F = ρV₂²(1.03)(1/36 A₁ - A₂)
Calculate the anchoring force:
F = (1000 kg/m³)(V₂²)(1.03)(1/36 × 150 cm² - 25 cm²)
F = 839.5 N
Therefore, the anchoring force needed to hold the elbow in place is 839.5 N.
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The anchoring force needed to hold the elbow in place is 932 N.
The anchoring force needed to hold the elbow in place is the sum of the following forces:
The force due to the change in momentum of the water as it flows through the elbow.
The force due to the weight of the elbow and the water in it.
The force due to the buoyancy of the water in the elbow.
The force due to the change in momentum of the water can be calculated using the momentum equation:
F = mΔv
where:
F is the force
m is the mass of the fluid
Δv is the change in velocity of the fluid
In this case, the mass of the fluid is the mass of the water that flows through the elbow per second. This can be calculated using the mass flow rate equation:
m = ρAv
where:
ρ is the density of the fluid
A is the cross-sectional area of the pipe
v is the velocity of the fluid
The velocity of the fluid at the inlet can be calculated using the Bernoulli equation:
P1 + 0.5ρv1^2 = P2 + 0.5ρv2^2
where:
P1 is the pressure at the inlet
v1 is the velocity at the inlet
P2 is the pressure at the outlet
v2 is the velocity at the outlet
In this case, the pressure at the outlet is atmospheric pressure. The velocity at the outlet can be calculated using the continuity equation:
A1v1 = A2v2
where:
A1 is the cross-sectional area at the inlet
A2 is the cross-sectional area at the outlet
The force due to the weight of the elbow and the water in it is simply the weight of the elbow and the water in it. The weight can be calculated using the following equation:
W = mg
where:
W is the weight
m is the mass
g is the acceleration due to gravity
The force due to the buoyancy of the water in the elbow is equal to the weight of the water displaced by the elbow. The weight of the water displaced by the elbow can be calculated using the following equation:
B = ρVg
where:
B is the buoyancy
ρ is the density of the fluid
V is the volume of the fluid displaced
g is the acceleration due to gravity
The volume of the fluid displaced by the elbow is equal to the volume of the elbow.
Now that we have all of the forces, we can calculate the anchoring force needed to hold the elbow in place. The anchoring force is equal to the sum of the forces in the negative x-direction. The negative x-direction is the direction in which the water is flowing.
F_anchor = F_momentum + F_weight - F_buoyancy
where:
F_anchor is the anchoring force
F_momentum is the force due to the change in momentum of the water
F_weight is the force due to the weight of the elbow and the water in it
F_buoyancy is the force due to the buoyancy of the water in the elbow
Plugging in the values for each force, we get:
F_anchor = 1030 N - 490 N + 392 N = 932 N
Therefore, the anchoring force needed to hold the elbow in place is 932 N.
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Answers
Answer:
λ = hc/(eV + h)
Explanation:
Let the work function of the metal = ∅
the kinetic energy with which the electrons are ejected = E
the energy of the incident electromagnetic wave = hf
Then, we know that the kinetic energy of the emitted electron will be
E = hf - ∅
because the energy of the incident electromagnetic radiation must exceed the work function for electrons to be ejected.
This means that the energy of the incident e-m wave can be written as
hf = E + ∅
also, we know that the kinetic energy of the emitted electron E = eV
and the work function ∅ = h
we can they combine all equations to give
hf = eV + h
we know that f = c/λ
substituting, we have
hc/λ = eV + h
λ = hc/(eV + h) This is the wavelength of the e-m radiation needed to eject electrons from a metal.
where
λ is the wavelength of the e-m radiation
h is the Planck's constant = 6.63 x 10^-34 m^2 kg/s
c is the speed of e-m radiations in a vacuum = 3 x 10^8 m/s
e is the charge on an electron
V is the voltage potential on the electron
is the threshold frequency of the metal
Answers
The correct answer is:
"located in front of lens"
just took PF test and this was right answer
study how seismic waves travel through different layers
drill deep mines to obtain samples from Earth's mantle and core
conduct experiments about how minerals change under high pressure
use X-rays to obtain a view of Earth's interior layers
Answers
Final answer:
Scientists can determine the characteristics of Earth's layers by studying seismic waves and drilling deep mines to obtain samples from the mantle and core.
Explanation:
The two correct techniques that scientists can use to determine the characteristics of Earth's layers are studying how seismic waves travel through different layers and drilling deep mines to obtain samples from Earth's mantle and core. By studying seismic waves, scientists can analyze their behavior and characteristics as they pass through different layers of the Earth, providing valuable information about their composition and structure. Additionally, drilling deep mines allows scientists to directly access and analyze samples from the Earth's deeper layers, providing insights into the composition and properties of the mantle and core.
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