Ask question

Ask question

All categories

3 years ago

15

The displacement vector from your house to the library is 520 m long, pointing 40 ∘ north of east.What are the x-component (x-ax

is is directed from west to east) of this displacement vector?What are the y-component (y-axis is directed from south to north) of this displacement vector?

1 answer:

Alexeev081 [22]3 years ago

7 0

Answer:

398.3 m, 334.2 m

Explanation:

The magnitude of the displacement vector is

v = 520 m

And its direction is

$\theta=40^{\circ}$

measured as north of east.

The x-component of this vector is given by:

$v_x = v_0 cos \theta = (520)cos 40^{\circ}=398.3 m$

While the y-component is given by

$v_y = v_0 sin \theta =(520)sin 40^{\circ}=334.2 m$

You might be interested in

Is work being done if you push a child in a wagon

Leto [7]

yes. youre using energy to make the wagon with the child in the wagon

6 0

4 years ago

Read 2 more answers

When does DNA replication occur?

Digiron [165]

Answer:

NA is replicated during the S phase (Synthesis phase) of Interphase.

Hope it helps!!!!!!!!!!

5 0

3 years ago

Read 2 more answers

A car with 2 × 10^3 kg moving at a speed of 10 m/s collides and sticks with car B of mass of 3 × 10^3 kg initially at rest. How

stepan [7]

Answer:

$6 \times 10^4 \; \rm J$ .

Explanation:

KE lost = Total KE before Collision - Total KE after Collision.

For each car, the KE before collision can simply be found with the equation:

$\displaystyle \mathrm{KE} = \frac{1}{2}\, m \cdot v^2$ , where

$m$ is the mass of the car, and
$v$ is the speed of the car.

The $2 \times 10^3\; \rm kg$ car would have an initial KE of:

$\displaystyle \frac{1}{2} \times 2 \times 10^3 \times 10^2 = 10^5\; \rm J$ .

The $3 \times 10^3\; \rm kg$ car was initially not moving. Hence, its speed and kinetic energy would zero before the collision.

To find the velocity of the two cars after the collision, apply the conservation of momentum.

The momentum $p$ of an object is equal to its mass $m$ times its velocity $v$ . In other words, $p = m\cdot v$ .

Let the mass of the two cars be denoted as $m_1$ and $m_2$ , and their initial speeds $v_1$ and $v_2$ . Since the two cars are stuck to each other after the collision, their final speeds would be the same. Let that speed be denotes as $v_3$ .

Initial momentum of the two-car system:

$\begin{aligned}& m_1 \cdot v_1 + m_2 \cdot v_2 \\ &= 2 \times 10^3 \times 10 + 3 \times 10^3 \times 0 \\ &= 2 \times 10^4\; \rm kg \cdot m \cdot s^{-1}\end{aligned}$ .

After the collision, both car would have a velocity of $v_3$ (since they were stuck to each other.) As a result, the final momentum of the two-car system would be:

$m_1\cdot v_3 + m_2 \cdot v_3 = (m_1 + m_2)\, v_3$ .

Since momentum is conserved during the collision, the momentum of the system after the collision would also be $2 \times 10^4 \; \rm kg \cdot m \cdot s^{-1}$ . That is: $(m_1 + m_2) \, v_3 = 2 \times 10^4 \; \rm kg \cdot m \cdot s^{-1}$ .

Solve for $v_3$ :

$\begin{aligned} v_3 &= \frac{(m_1 + m_2)\, v_3}{m_1 + m_2} \\ &= \frac{2 \times 10^4}{2 \times 10^3 + 3 \times 10^3} \\ &= \frac{2 \times 10^4}{5 \times 10^3} \\ &= 4 \; \rm m \cdot s^{-1}\end{aligned}$ .

Hence, the total kinetic energy after the collision would be:

$\begin{aligned} &\frac{1}{2}\, m_1 \, v^2 + \frac{1}{2}\, m_2\, v^2 \\ &= \frac{1}{2}\, (m_1 + m_2)\, v^2 \\ &= \frac{1}{2} \times \left(2 \times 10^3 + 3 \times 10^3\right) \times 4^2 \\ &= 4 \times 10^4\; \rm J\end{aligned}$ .

The amount of kinetic energy lost during the collision would be:

$\begin{aligned}&\text{KE After Collision} - \text{KE Before Collision} \\ &= 10^5 - 4 \times 10^4 \\&= 6\times 10^4\; \rm J \end{aligned}$ .

5 0

3 years ago

A motor-driven winch pulls a 50.0 kg student 5.00 m up the rope at a constant speed of 1.25 m/s. how much power does the motor u

nadya68 [22]

Power is the rate work done given by dividing work done by unit time. It is measured in watts equivalent to J/s.
In this case the force by the student is mg = 490 N (taking g as 9.8m/s²)
Work done is given by force × distance,
Therefore, Power =(force × distance)/ time, but velocity/speed =distance/time
Thus, Power = force × speed/velocity
= 490 N × 1.25
= 612.5 J/S (Watts)
Hence, power will be 612.5 Watts.

7 0

4 years ago

What does it mean if the slope is zero?

kiruha [24]

The object is not moving.

My explanation is that say if you sit a ball on the table and it is a smooth surface with no bumps or anything. The ball will sit still since it can’t roll unless you hit it.

Hope this helps!

5 0

3 years ago