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ValentinkaMS [17]

3 years ago

10

A projectile is launched horizontally from a height of 8.0 m. The projectile travels 6.5 m before hitting the ground.

1 answer:

spayn [35]3 years ago

7 0

Answer:

5.09 m/s

Explanation:

Use the height to find the time it takes to land:

y = y₀ + v₀ᵧ t + ½ gt²

0 = 8.0 m + (0 m/s) t + ½ (-9.8 m/s²) t²

t = 1.28 s

Now use the horizontal distance to find the initial velocity.

x = x₀ + v₀ₓ t + ½ at²

6.5 m = 0 m + v₀ (1.28 s) + ½ (0 m/s²) (1.28 s)²

v₀ = 5.09 m/s

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What is the average velocity of a wave that travels an average distance of 6 m in 0.25 s?

Dmitriy789 [7]

S=Vt
V=S/t
V= 6/0.25
V=24m/s

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3 years ago

Describe two ways to change the frictional force between two solid surfaces.

Tanya [424]

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4 years ago

A motorist runs out of gas on a level road 230m from a gas station. The driver pushes the 1140kg car to the gas station. If a 14

malfutka [58]

Answer:

The work done is $32.2kJ$

Explanation:

Work is defined as the product of force and distance moved in the direction of application of force.

$W= F*S$

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force F= 140 N

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The work done by pushing the car for a distance of $230m$ is $32.2kJ$

7 0

3 years ago

The rotational inertia of a thin rod about one end is 1/3 ML2. What is the rotational inertia of the same rod about a point loca

zlopas [31]

Answer:

The value is $I = 0.0932 ML ^2$

Explanation:

From the question we are told that

The rotational inertia about one end is $I_R = \frac{1}{3} ML^2$

The location of the axis of rotation considered is $d = 0.4 L$

Generally the mass of the portion of the rod from the axis of rotation considered to the end of the rod is $0.4 M$

Generally the length of the rod from the its beginning to the axis of rotation consider is

$k = 1 - 0.4 L = 0.6L$

Generally the mass of the portion of the rod from the its beginning to the axis of rotation consider is

$m = 1- 0.4 M = 0.6 M$

Generally the rotational inertia about the axis of rotation consider for the first portion of the rod is

$I_{R1} = \frac{1}{3} (0.6 M )(0.6L)^2$

$I_{R1} = \frac{1}{3} (0.6 M )L^2 0.6^2$

Generally the rotational inertia about the axis of rotation consider for the second portion of the rod is

$I_{R2} = \frac{1}{3} (0.6 M )(0.6L)^2$

=> $I_{R2} = \frac{1}{3} (0.4 M )(0.4L)^2$

=> $I_{R2} = \frac{1}{3} (0.4 M )L^2 0.4^2$

Generally by the principle of superposition that rotational inertia of the rod at the considered axis of rotation is

$I = \frac{1}{3} (0.6 M )L^2 0.6^2 + \frac{1}{3} (0.4 M )L^2 0.4^2$

=> $I = \frac{1}{3} ML ^2 [0.6 * (0.6)^2 + 0.4 * (0.4)^2 ]$

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8 0

3 years ago

What is the power of a kitchen blender if it can perform 2000 joules of work in 5 seconds ?

soldier1979 [14.2K]

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3 years ago