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r-ruslan [8.4K]
3 years ago
11

State True or False:If the positive x-axis points to the right and the x-component of velocity is negative, the cart must be mov

ing to the left.
Physics
1 answer:
umka2103 [35]3 years ago
7 0

Answer:

True

Explanation:

Velocity is a vector, therefore it consists of two elements:

- A magnitude (also called speed), which is the ratio between the displacement of the object and the time taken

- A direction, which corresponds to the direction of motion of the object

Therefore, velocity can be describes as positive or negative, depending on the direction which has been chosen as positive. If we chose the positive x-axis as positive direction, therefore:

- if the object is moving to the right (positive x-direction), the velocity will be positive

- if the object is moving to the left (negative x-direction), the velocity will be negative

So, in this case, since the velocity of the cart is negative, it must be moving to the left.

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What are electromagnetic waves?
adelina 88 [10]

D: Sóng có thể truyền qua khoảng không của không gian,

với tốc độ ánh sáng.

3 0
3 years ago
How many significant figures?<br> 5.0001<br> O None of these are correct<br> O 5<br> 02<br> 0 1
mezya [45]

5

if zero falls between two significant numbers it becomes significant.

6 0
3 years ago
What the unit of work?​
ra1l [238]

Answer:

yes

Explanation:

5 0
2 years ago
An electron with a speed of 1.7 × 107 m/s moves horizontally into a region where a constant vertical force of 4.9 × 10-16 N acts
Alex Ar [27]

Answer:

 y = 77.74 10⁻⁵ m

Explanation:

For this exercise we can use Newton's second law

        F = m a

        a = F / m

        a = 4.9 10⁻¹⁶ / 9.1 10⁻³¹

        a = 0.538 10¹⁵ m / s

This is the vertical acceleration of the electron.

Now let's use kinematics to find the time it takes to move the

         x= 29 mm = 29 10⁻³ m

On the x axis

            v = x / t

            t = x / v

            t = 29 10⁻³ / 1.7 10⁷

            t = 17 10⁻¹⁰ s

Now we can look for vertical distance at this time.

            y = v_{oy} t + ½ a t²

            y = 0 + ½ 0.538 10¹⁵ (17 10⁻¹⁰)²

            y = 77.74 10⁻⁵ m

3 0
3 years ago
Read 2 more answers
What happens to the force between two charges when each charge is doubled and the distance between them is 1/4 its original
DIA [1.3K]

Answer:

F' = 64 F

Explanation:

The electric force between charges is given by :

F=\dfrac{kq_1q_2}{r^2}

Where

q₁ and q₂ are charges

r is the distance between charges

When  each charge is doubled and the distance between them is 1/4 its original magnitude such that,

q₁' = 2q₁, q₂' = 2q₂ and r' = (r/4)

New force,

F'=\dfrac{kq_1'q_2'}{r'^2}

Apply new values,

F'=\dfrac{k\times 2q_1\times 2q_2}{(\dfrac{r}{4})^2}\\\\=\dfrac{k\times 4q_1q_2}{\dfrac{r^2}{16}}\\\\=64\times \dfrac{kq_1q_2}{r^2}\\\\=64F

So, the new force becomes 64 times the initial force.

7 0
2 years ago
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