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GenaCL600 [577]
3 years ago
8

A glider is gliding through the air at a height of 416 meters with a speed of 45.2 m/s. The glider

Physics
1 answer:
julia-pushkina [17]3 years ago
3 0

There's not enough information to find an answer.

I think the idea here is that in descending (416 - 278) = 138 meters,
the glider gives up some gravitational potential energy, which
becomes kinetic energy at the lower altitude.  This is all well and
good, but we can't calculate the difference in potential energy
without knowing the mass of the glider.

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A 375-pound concrete cylinder has a base area of 144 square inches. with the cylinder resting on its base, the pressure exerted
Akimi4 [234]

The pressure exerted by the concrete cylinder is 2.60 pound/in².

We need to know about the pressure to solve this problem. Pressure is a unit that describes how much force is applied to a surface area. It can be determined as

P = F / A

where P is pressure, F is force and A is area.

From the question above, we know that

F = 375 pound

A = 144 in²

By substituting the given parameters, we can calculate the pressure

P = F / A

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P = 2.60 pound/in²

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6 0
1 year ago
Vector vector a has a magnitude of 29 units and points in the positive y-direction. when vector vector b is added to vector a ,
Nutka1998 [239]
Good morning.

We have:

\mathsf{\overset{\to}{a} = 29\overset{\to}{j}}

Where j is the unitary vector in the direction of the y-axis.

We have that 

\mathsf{\overset{\to}{a}+\overset{\to}{b} = -18\overset{\to}{j}}

We add the vector -a to both sides:

\mathsf{\overset{\to}{b} = -18\overset{\to}{j} -\overset{\to}{a} = -18\overset{\to}{j} -29\overset{\to}{j}}\\ \\ \mathsf{\overset{\to}{b}=-47\overset{\to}{j}}


Therefore, the magnitude of b is 47 units.
5 0
2 years ago
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A 35.9 g mass is attached to a horizontal spring with a spring constant of 18.4 N/m and released from rest with an amplitude of
lidiya [134]

Answer:

7.74m/s

Explanation:

Mass = 35.9g = 0.0359kg

A = 39.5cm = 0.395m

K = 18.4N/m

At equilibrium position, there's total conservation of energy.

Total energy = kinetic energy + potential energy

Total Energy = K.E + P.E

½KA² = ½mv² + ½kx²

½KA² = ½(mv² + kx²)

KA² = mv² + kx²

Collect like terms

KA² - Kx² = mv²

K(A² - x²) = mv²

V² = k/m (A² - x²)

V = √(K/m (A² - x²) )

note x = ½A

V = √(k/m (A² - (½A)²)

V = √(k/m (A² - A²/4))

Resolve the fraction between A.

V = √(¾. K/m. A² )

V = √(¾ * (18.4/0.0359)*(0.395)²)

V = √(0.75 * 512.53 * 0.156)

V = √(59.966)

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Find the velocity and distance travelled of a car that is accelerating at 3 m/s2 for 4 seconds.
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Explanation:

Given:

v₀ = 0 m/s

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Find: Δx and v

Δx = v₀ t + ½ at²

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Δx = 24 m

v = at + v₀

v = (3 m/s²) (4 s) + 0 m/s

v = 12 m/s

6 0
2 years ago
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