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1 month ago

If you double the mass of an object while leaving the net force unchanged what is the result

Physics

1 answer:

kicyunya [3.2K]1 month ago

5 0

Answer:

If the net force acting on an object doubles, its acceleration also doubles. Conversely, if the mass is doubled, the acceleration will be halved. If both the net force and mass are doubled, the acceleration remains the same.

Explanation:

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A fire hose held near the ground shoots water at a speed of 6.5 m/s. At what angle(s) should the nozzle point in order that the

inna [3103]

The velocity of water can be decomposed into its vertical and horizontal components:
$v_x = 6.5 cos \theta \\ v_y = 6.5 sin \theta$

The vertical component will exhibit a parabolic trajectory due to gravity, while the horizontal component will be linear:
$y(t) = -4.9t^2 + (6.5sin \theta) t \\ \\ x(t) = (6.5 cos \theta) t$
To determine when the water reaches the ground 2.5m away, set y= 0 and x = 2.5
$-4.9t^2 + (6.5sin \theta) t=0 \\ \\ t = \frac{6.5}{4.9} sin \theta \\ \\(6.5 cos \theta)(\frac{6.5}{4.9} sin \theta) = 2.5 \\ \\ sin \theta cos \theta = 0.29 \\ \\ sin 2\theta = 0.58 \\ \\ 2\theta = 35.4, 144.6 \\ \\ \theta = 17.7,72.3$

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2 months ago

An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0 km/hour is blowing southward.

Softa [3030]

Answer:

The plane's speed in relation to the ground is 300.79 km/h.

Explanation:

Provided details include:

Wind speed = 75.0 km/hr

Plane's airspeed = 310 km/hr

Next, we must find the ground speed of the plane

Calculating the angle

Using the angle formula

$\sin\theta=\dfrac{v'}{v}$

Where v' represents the wind speed

v represents the plane's speed

We will substitute the values into the formula

$\sin\theta=\dfrac{75}{310}$

$\theta=\sin^{-1}(\dfrac{75}{310})$

$\theta=14.0^{\circ}$

Now, we must find the resultant speed

Using the resultant speed formula

$\cos\theta=\dfrac{v''}{v}$

Insert the values into the formula

$\cos14=\dfrac{v''}{310}$

$v''=\cos14\times310$

$v''=300.79\ km/h$

Consequently, the plane's speed in relation to the ground equals 300.79 km/h.

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2 months ago

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Maru [3345]

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1 month ago

A 0.050-kg lump of clay moving horizontally at 12 m/s strikes and sticks to a stationary 0.15-kg cart that can move on a frictio

Ostrovityanka [3204]

Answer:

Explanation:

According to the parameters provided,

mass of the clay lump, m₁ = 0.05 kg

initial velocity of the lump, u₁ = 12 m/s

mass of the cart, m₂ = 0.15 kg

initial speed of the cart, u₂ = 0

As the clay adheres to the cart, we have an inelastic collision scenario. Let v represent the combined speed of both the cart and lump post-collision. Given that momentum is conserved, we have:

$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$v=\dfrac{0.05\ kg\times 12\ m/s+0}{0.05\ kg+0.15\ kg}$

The resultant speed is v = 3 m/s.

Thus, the final speed of both cart and lump following the collision is 3 m/s. This concludes the solution.

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2 months ago