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2 years ago

Sand dunes during the Dust Bowl are an example of:

Physics

1 answer:

choli [55]2 years ago

5 0

Sand dunes during the Dust Bowl are an example of windward.

The sand dunes are formed due to high movement of wind in the deserts which make dunes in the direction where wind is moving. Windward refers to facing the wind or situated on the side facing the wind so in those regions where the wind moves bring sand with itself and make dunes in that area so we can conclude that sand dunes are formed in the windward regions.

Learn more: brainly.com/question/24879138

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An object has a mass of 785 g and a volume of 15 cm³. What is its density? (Give your answer in g/cm³ to 2 decimal places).

Anvisha [2.4K]

Answer:

denisity = 52.33 g/c $m^{3}$

Explanation:

Density:

$d = \frac{m}{v}$

We have that m = 785 and that v = 15 c $m^{3}$ .

$d = \frac{785}{15}$

d = 52.33 $m^{3}$

4 0

2 years ago

An object is hanging by a string from the ceiling of an elevator. The elevator is moving upward with a constant speed. What is t

lana [24]

Answer:

Equal to the weight of the object

Explanation:

There are two forces acting on the object hanging on the string:

- The tension in the string, T, acting upward

- The weight of the object, mg, acting downward

We can write Newton's second law as:

$T-mg=ma$

where a is the acceleration of the object.

Here we are told that the elevator is moving at constant speed: this means that the acceleration of the object is zero, therefore a = 0 and the equation becomes

$T=mg$

therefore, the tension in the string is equal to the weight of the object.

7 0

3 years ago

Dos coches, uno de 1.000 kg que circula a 100 km/h, y otro de 1.200 kg que circula a 90

Kobotan [32]

The collision of car 2 is more violent (because more impulse is exerted)

Explanation:

The collision which is more violent is the one in which more impulse is exchanged.

The impulse exerted by each car on the wall is equal to the change in momentum of the car:

$I=m\Delta v$

where

m is the mass of the car

$\Delta v$ is the change in velocity

For the car 1,

m = 1000 kg

$\Delta v = -100 km/h \cdot \frac{1000 m/km}{3600 s/h}=-27.8 m/s$ (the sign is negative because the velocity of the car has changed from 100 km/h to 0 km/h)

So the magnitude of the impulse of car 1 is

$I_1 = (1000)(27.8)=27,800 kg m/s$

For the car 2,

m = 1200 kg

$\Delta v = -90 km/h \cdot \frac{1000 m/km}{3600 s/h}=-25 m/s$

So the magnitude of the impulse of car 2 is

$I_2 = (1200)(25)=30,000 kg m/s$

So, car 2 exerts a larger impulse, therefore its impact is more violent.

Learn more about impulse and change in momentum:

brainly.com/question/9484203

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7 0

2 years ago

wire 1 and 2 are made of the same metal. wire 2 has twice the length and diameter of the of wire 1.what is the ratio R2/R1 of th

Vlada [557]

Since the wires are made up of the same material, the resistivities will be the same.

4 0

2 years ago

Two cars collide at an intersection. Car A , with a mass of 2000kg , is going from west to east, while car B , of mass 1400kg ,

ahrayia [7]

Complete Question:

Two cars collide at an intersection. Car A , with a mass of 2000 kg, is going from west to east, while car B, of mass 1500 kg, is going from north to south at 15 m/s. As a result, the two cars become enmeshed and move as one. As an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of 65∘ south of east from the point of impact. (a) How fast were the enmeshed cars moving just after the collision? (b) How fast was car A going just before the collision?

Answer:

a) 6.36 m/s b) 4.57 m/s

Explanation:

a) Assuming no external forces acting during the collision, total momentum must be conserved.

As momentum is a vector, we can decompose it along two directions perpendicular each other.

Just for convenience, we choose as our x-axis to the W-E direction, and as our y-axis, the direction N-S.

If we know that total momentum must be conserved, same must be true for both components, px and py.

Applying the information provided (both cars become enmeshed after the collision, moving at an angle of 65º south of east from the point of impact), we have:

px = ma * va = (ma+mb) * vab * cos 65º (1)

py = mb * vb = (ma + mb) * vab * sin 65º (2)

Replacing by the values of ma, mb, and sin 65º, we can solve for vab, as follows:

vab = 1,400 kg* 14.0 m/s / (3,400 Kg * sin 65º) = 6.36 m/s

b) Replacing vab from above in (1), and solving for va, we have:

va = 3,400 kg* 6.36 m/s* cos 65º / 2,000 Kg = 4.57 m/s

7 0

2 years ago