All categories

3 years ago

____________ is a medium-sized flat-topped hill with cliff-face sides that is taller than it is wide.

Physics

1 answer:

Luda [366]3 years ago

5 0

Answer:

mesa

Explanation:

A mesa is a flat-topped mountain or hill. It is a wide, flat, elevated landform with steep sides. ... Spanish explorers of the American southwest, where many mesas are found, used the word because the tops of mesas look like the tops of tables.

You might be interested in

Rachel and Jason were pulling on a remote control from opposite ends. If Rachel was pulling west with a force of 100 N and the n

juin [17]

-Jason would be pulling with a force of 60N from east, as east is opposite from west, and if the remote control is 40N, and if Rachel is using a force of 100N - 40N would be 60N east.

-Hope this helps! xo

8 0

3 years ago

A 465 g block slides along a frictionless surface at a speed of 0.35 m/s. It runs into a horizontal massless spring with spring

katovenus [111]

Answer:

a) The duration, during which the block remain in contact with the spring is 0.29 s

b) The period of the simple harmonic oscillatory motion depends only on the mass and spring constant, therefore when the speed is doubled, the duration of contact remains the same as 0.29 s.

Explanation:

Mass of the block = 465 g

Surface speed = 0.35 m/s

Spring constant , k = 54 N/m

$T = 2\times \pi \times \sqrt{\frac{m}{K} } = 2\times \pi \times \sqrt{\frac{0.465}{54} }$ = 0.58 s

a) Since the period for the oscillatory motion is 0.58 s, then the time when the block and spring remain in contact is T/2 = 0.29 s

b) When the speed is doubled, we have

$T = 2\times \pi \times \sqrt{\frac{m}{K} }$

Therefore, since T is only dependent on the mass, m and the spring constant, K, then the time it takes when the speed is doubled remain as

T /2 = 0.29 s

7 0

3 years ago

An electron that has a velocity with x component 2.4 x 106 m/s and y component 3.6 x 106 m/s moves through a uniform magnetic fi

likoan [24]

Answer:

(a) 7.315 x 10^(-14) N

(b) - 7.315 x 10^(-14) N

Explanation:

As you referred at the final remark, the electron and proton undergo a magnetic force of same magnitude but opposite direction. Using the definition of magnetic force, a cross product must be done. One technique is either calculate the magnitude of the velocity and magnetic field and multiplying by sin (90°), but it is necessary to assure both vectors are perpendicular between each other ( which is not the case) or do directly the cross product dealing with a determinant (which is the most convenient approach), thus,

(a) The electron has a velocity defined as: $\overrightarrow{v}=(2.4x10^{6} i + 3.6x10^{6} j) \frac{[m]}{[s]}\\\\$

In respect to the magnetic field; $\overrightarrow{B}=(0.027 i - 0.15 j) [T]$

The magnetic force can be written as;

$\overrightarrow{F} = q(\overrightarrow{v} x \overrightarrow{B})\\ \\\\\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]$

Bear in mind $q =-1.6021x10^{-19} [C]$

thus,

$\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]\\\\\\\overrightarrow{F}= q(2.4x10^{6}* (-0.15)- (0.027*3.6x10^{6}))\\\\\\\overrightarrow{F}= -1.6021x10^{-19} [C](-457200) [T]\frac{m}{s}\\\\\overrightarrow{F}=(7.3152x10^{-14}) k [\frac{N*m/s}{C*m/s}]\\\\|F|= \sqrt{ (7.3152x10^{-14})^{2}[N]^2 *k^{2}}\\\\F=7.3152x10^{-14} [N]$

Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e, $k^{2}=k\cdot k = 1$

(b) Considering the proton charge has the same magnitude as electron does, but the sign is positive, thus

$\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]\\\\\\\overrightarrow{F}= q(2.4x10^{6}* (-0.15)- (0.027*3.6x10^{6}))\\\\\\\overrightarrow{F}= 1.6021x10^{-19} [C](-457200) [T]\frac{m}{s}\\\\\overrightarrow{F}=(-7.3152x10^{-14}) k [\frac{N*m/s}{C*m/s}]\\\\|F|= \sqrt{ (-7.3152x10^{-14})^{2}[N]^2 *k^{2}}\\\\F=-7.3152x10^{-14} [N]$

Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e, $k^{2}=k\cdot k = 1$

Final remarks: The cross product was performed in R3 due to the geometrical conditions of the problem.

6 0

4 years ago

When hydrogen reacts with oxygen, hydrogen loses electrons. Which term

Tanya [424]

Answer: Correct answer is A

Explanation: Ap3x appointment

4 0

3 years ago

Read 2 more answers

Why does it rain more in West Ferris than in East Ferris? Explain your answer.

daser333 [38]

Answer:

This idea helps students explain why more rain forms over West Ferris than East Ferris. ... Therefore, when students explain that water vapor condenses higher in the atmosphere, they are actually explaining that water vapor condenses high in the troposphere, which is relatively low in the atmosphere.

Explanation:

Plz mark me brainliest thank u> have a good day

7 0

3 years ago