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Snezhnost [94]
4 years ago
12

Suppose the stone is thrown at an angle of 39.0° below the horizontal from the same building as in the Example above. If it stri

kes the ground 47.8 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.)
(a) the time of flight
s
The x coordinate as a function of time is x(t) = vcos(39.0)t, so the initial speed is v0 = Δx/(cos 39.0Δt), where Δx = 47.8 and Δt is the time of flight. Insert this into your equation for y(t) and solve for the time of flight.
Physics
1 answer:
Anni [7]4 years ago
5 0
Suppose the stone is thrown at an angle of 39.0° below the horizontal from the same building as in the Example above. If it strikes the ground 47.8 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.)(a) the time of flight  sThe x coordinate as a function of time is x(t) = vcos(39.0)t, so the initial speed is v0 = Δx/(cos 39.0Δt), where Δx = 47.8 and Δt is the time of flight. Insert this into your equation for y(t) and solve for the time of flight. Note that the answer should be smaller than 3.16227766016838, since the stone is thrown down (and to the right).(b) the initial speed  m/s(c) the speed and angle of the velocity vector with respect to the horizontal at impactspeed  m/sangle °
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Un puente(construido de acero) que conduce hacia una bahía mide 1.30X10-3m entre sus torres de apoyo. ¿Cuánto se dilata el puent
JulsSmile [24]

Answer:

ΔL = 0.429 m

Explanation:

The expression for linear thermal expansion is

         ΔL = α L₀ ΔT

the coefficient of linear expansion of steel is 11 10⁻⁶ ºC⁻¹

the initial length of the bridge is L₀ = 1.30 10³ m and the temperature variation is ΔT = 30 ºC

calculate  

        ΔL = 11 10⁻⁶  1.30 10³  30

        ΔL = 4.29 10⁻¹ m

        ΔL = 0.429 m

8 0
3 years ago
A sound wave of wavelength 0.682 m and velocity 325 m/s is produced for 0.500 s. what is the frequency
Naya [18.7K]
Recall that given the velocity and wavelength, frequency can be computed as

frequency =  \frac{speed}{wavelength}
Substituting the given values, we have
frequency =  \frac{325 m/s}{0.682 m}  \\ frequency = 477 Hz

Answer: 477 Hz
6 0
3 years ago
Determine the total amount of heat, in joules, required to completely vaporize a 50.0-gram sample of h2o(?) at its boiling point
Tanzania [10]
In order to calculate the amount of energy required, we must first check the latent heat of vaporization of water from literature. The latent heat of vaporization of any substance is the amount of energy required per unit mass to convert that substance from a solid to a liquid. For water this is 2,260 J/g. We now use the formula:
Energy = mass * latent heat
Q = 50 * 2,260
Q = 113,000 J

113,000 Joules of heat energy are required.
3 0
3 years ago
An 89.2-kg person with a density 1025 kg/m3 stands on a scale while completely submerged in water. What does the scale read?
Luden [163]

Answer:

89.11kg

Explanation:

Note an object weighs less when in a fluid and the weight of the volume of the fluid displaced is known as the upthrust.

Now, the person is going to displace the volume 89/1025 =0.087m3 { from density D = mass(M)/volume(V)}

The weight of the fluid displaced is the density of the fluid × volume of fluid displaced.

The weight of the fluid=0.087m3× 1kg/me = 0.087kg

Now the weight of the fluid displaced is referred to as the upthrust.

Now the real weight - the apparent weight = the upthrust.

Hence the apparent weight = real weight - upthrust

Apparent weight = 89.2-0.087 = 89.11kg

3 0
3 years ago
A 0.0600-kilogram ball traveling at 60.0 meters per second hits a concrete wall. What speed must a 0.0100-kilogram bullet have i
GarryVolchara [31]

Answer:

the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is 360 m/s

Explanation:

The computation of the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is as follows:

The ball momentum is

\Delta Q = mv\\\\\Delta Q = 6 \times 1^-2 \times 60\\\\\Delta Q = 3.6 kg \times m/s

As there is a similar momentum

So,

\Delta Q = MV\\\\3.6 = 10^-2V\\\\V = 360 m/s

Hence, the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is 360 m/s

5 0
3 years ago
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