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luda_lava [24]
2 years ago
5

A person drives a car for a distance of 450.0 m. The displacement A of the car is illustrated in the drawing. What are the scala

r components of this displacement vector?
Physics
1 answer:
dedylja [7]2 years ago
5 0

Answer:

Scalar quantity is the distance travelled i.e 450 meters

Explanation:

The scalar component of displacement is the distance travelled which is 450 meters.

The vector component is basically the direction of the displacement vector.

Hence, scalar quantity is the distance travelled i.e 450 meters

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A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit
inn [45]

Answer:twice of initial value

Explanation:

Given

spring compresses x_1 distance for some initial speed

Suppose v is the initial speed and k be the spring constant

Applying conservation of energy

kinetic energy converted into spring Elastic potential energy

\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_1^2----1

When speed doubles

\dfrac{1}{2}m(2v)^2=\dfrac{1}{2}kx_2^2----2

divide 1 and 2

\dfrac{1}{4}=\dfrac{x_1^2}{x_2^2}

x_2=2x_1

Therefore spring compresses twice the initial value

   

7 0
3 years ago
A brick lands 10.1 m from the base of a building. If it was given an initial velocity of 8.6 m/s [61º above the horizontal], how
Montano1993 [528]
<h2>Answer: 10.52m</h2><h2 />

First, we have to establish the <u>reference system</u>. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).

According to this, the initial velocity V_{o} has two components, because the brick was thrown at an angle \alpha=61\º:

V_{ox}=V_{o}cos\alpha   (1)

V_{ox}=8.6\frac{m}{s}cos(61\º)=4.169\frac{m}{s}  (2)

V_{oy}=V_{o}sin\alpha   (3)

V_{oy}=8.6\frac{m}{s}sin(61\º)=7.521\frac{m}{s}   (4)

As this is a projectile motion, we have two principal equations related:

<h2>In the x-axis: </h2>

X=V_{ox}.t  (5)

Where:

X=10.1m is the distance where the brick landed

t is the time in seconds

If we already know X and V_{ox}, we have to find the time (we will need it for the following equation):

t= \frac{X}{ V_{ox}}  (6)

t=2.42s  (7)

<h2>In the y-axis: </h2>

-y=V_{oy}.t+\frac{1}{2}g.t^{2}   (8)

Where:

y is the height of the building (<u>in this case it has a negative sign because of the reference system we chose)</u>

g=-9.8\frac{m}{s^{2}} is the acceleration due gravity

Substituting the known values, including the time we found on equation (7) in equation (8), we will find the height of the building:

-y=(7.521\frac{m}{s})(2.42s)+\frac{1}{2}(-9.8\frac{m}{s^{2}}).(2.42s)^{2}   (9)

-y=-10.52m   (10)

Multiplying by -1 each side of the equation:

y=10.52m >>>>This is the height of the building

3 0
3 years ago
Please need help on this one
jolli1 [7]

Answer:

The second answer from the top, no the energy in the wave pushed the water particles from above the earthquake in the opposite direction.

Explanation:

I believe this is the correct answer. Hope you do well

4 0
3 years ago
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Answer:

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Explanation:

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