Can someone please help me answer my question I’m really confused and I have a test tomorrow please and thank you god bless

A man walks 30 m to the west, then 5 m to the east in 45 seconds.

katen-ka-za [31]

Answer:

a. Displacement=30²+5²=925= 30.4m

b. Total distance=30m+5m=35m

c. V=s/t. = 30.4/45=0.6m/s

8 0

1 year ago

What is the magnetic force on a proton that is moving at 5.2 x 107 m/s to the

alisha [4.7K]

Answer:

1.1648×10⁻¹¹ N

Explanation:

Using

F = qvBsinФ..................... Equation 1

Where F = Force on the proton, q = charge, v = velocity, B = magnetic Field, Ф = angle between the magnetic Field and the velocity.

Note: The angle between v and B = 90°

Given: v = 5.2×10⁷ m/s, B = 1.4 T, q = 1.6×10⁻¹⁹ C, Ф = 90°

Substitute into equation 1

F = 1.6×10⁻¹⁹(5.2×10⁷)(1.4)sin90°

F = 11.648×10⁻¹²

F = 1.1648×10⁻¹¹ N.

6 0

3 years ago

Read 2 more answers

Question 8 of 10

-BARSIC- [3]

Answer:

The correct answer is the Convex lens. An image is formed when a ray of light coming from a point intersects at another point. The image is formed by the real intersection of light. The image is formed by the virtual intersection of Light.

here is the site : textbook.com

3 0

2 years ago

Object A of mass 0.70 kg travels horizontally on a frictionless surface at 20 m/s. It collides with object B, which is initially

AnnZ [28]

Answer:

25.71 kgm/s

Explanation:

Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.

Given that K₂ = 0.7K₁

1/2mv₂² = 0.7(1/2mv₁²)

v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s

Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.

Th magnitude of object A's momentum change is thus 25.71 kgm/s

6 0

3 years ago

A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. What is

Aleksandr [31]

Answer:

The acceleration of the crate is 1.8 m/s² so the answer is a.

Explanation:

The very first thing you must do when solving this problem is to draw a free body diagram. (The body diagram is attached to this answer)

So once we got the free body diagram, we can analyze it and build our sum of forces in the x and y directions. Notice that according to the diagram, there are 4 forces to this problem, Normal (N), Weight (W), kinetic friction (fk) and the 750N force.

As one may see in the free body diagram, two of the forces are vertical forces: N and W, so we can use them to build a sum of forces:

Starting with the sum of forces in the y-direction, we get:

Σ $F_{y}$ =0

We set the sum equal to zero because there is no movement in the y-direction, so the system is in vertical equilibrium.

so the sum will be:

$N-W=0$

when solving for N we get that:

N=W

where W is found by multiplying the mass of the crate by the acceleration of gravity:

N=250kg*9.8m/s²

N=2450N

Once we found the normal force, we can use it to find the kinetic friction which is given by the following formula:

$f_{k}=N$ μ

where μ is the kinetic friction coefficient.

So we get that the kinetic friction is:

$f_{k}=2450N*0.12$

so

$f_{k}=294$

With this information we can go ahead and find the sum of horizontal forces:

Σ $F_{x}$ =ma

In this case the sum is equal to mass times acceleration because the crate is moving horizontally due to the action of a force, so it will have an acceleration.

so the sum of forces look like this:

750N- $f_{k}$ =ma

so

750N-294N=(250kg)a

when solving for a we get:

$a=\frac{759N-294N}{250kg}\\ \\a=1.8m/s^{2}$

so the crate's acceleration is 1.82m/s².

5 0

3 years ago