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If an object accelerates at 10 m/s2 for 4 seconds, how much will its velocity change by?

jek_recluse [69]

Answer:

40 m/s

Explanation:

Acceleration formula: $a=\frac{\triangle v}{t}$

Δv=at=10*4=40

5 0

3 years ago

A kayaker paddles at 4.0 m/s in a direction 30° south of west. He then turns and paddles at 3.7 m/s in a direction 20° west of s

user100 [1]

7.2 m/s

49 degrees south of west

7 0

3 years ago

Read 2 more answers

At a certain location, a gravitational force with a magnitude of 350 newtons acts on a 70.-kilogram astronaut. What is the magni

creativ13 [48]

Answer:

3. 5.0N/kg

Explanation:

Gravitational field strength = gravitational force/mass of astronaut = 350N/70kg = 5.0N/kg

6 0

3 years ago

The force diagram shown here describes the forces that external objects (the surface andEarth) exert on a woman (in this scenari

sveticcg [70]

The three different motions are;

The upward motion of the woman is constant
The downward motion of the woman is also constant
The horizontal motion of the woman is zero.

<h3>What is force diagram?</h3>

Force diagram is a pictorial or graphical illustration of different forces acting on object.

In this given question, there two forces acting on the woman as depicted in the force diagram.

The first force is surface force (Fs)
The second force is force of Earth (FE)

In the given force diagram, the woman is in equilibrium, this implies that the surface force and the Earth force are equal.

The three different types of motion of the woman that are consistent with the force diagram include the following;

The upward motion of the woman is constant
The downward motion of the woman is also constant
The horizontal motion of the woman is zero since there is no horizontal force on the woman.

Learn more about force diagram here: brainly.com/question/3624253

#SPJ1

4 0

11 months ago

Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a

34kurt

Answer:

E= $\frac{KQ}{2\sqrt 2a^2}$

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

$E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}$

Substitute x=a and R=a

Then, we get

$E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{2\sqrt 2a^3}$

$E=\frac{KQ}{2\sqrt 2a^2}$

Where K= $9\times 10^9 Nm^2/C^2$

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center= $\frac{KQ}{2\sqrt 2a^2}$

4 0

3 years ago