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3 years ago

If you accelerate from rest at 10 m/s^2 what will be your speed after 6s

1 answer:

5 0

Ok so we know:
The acceleration (10ms^-2)
The initial velocity (0)
And the total time (6seconds)

Using the equation v = u + at, with v being the final velocity, u being the initial velocity, a being acceleration and t being time, we get:

v = 0 + 10x6
v = 60

So the speed after 6 seconds is 60 metres per second (6m/s)

Hope this helped

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Can u show me were these go on the picture:

PIT_PIT [208]

Answer:

Here u go

Explanation:

7 0

3 years ago

Ask Your Teacher Two long, straight wires are parallel and 11 cm apart. One carries a current of 2.9 A, the other a current of 5

dsp73

Answer:

The magnitude of force per unit length of one wire on the other is $2.7945\times 10^{-5}\ N$ and the direction is away from one another

The magnitude of force per unit length of one wire on the other is $2.7945\times 10^{-5}\ N$ and the direction is towards each other.

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi\times 10^{-7}\ N/A^2$

$i_1$ = Current in first wire = 2.9 A

$i_2$ = Current in second wire = 5.3 A

r = Gap between the wires = 11 cm

Force per unit length

$F_{12}=F_{21}=\frac{\mu_0 i_1i_2}{2\pi r}\\ =\frac{4\pi\times 10^{-7}\times 2.9\times 5.3}{2\pi 0.11}\\ =2.7945\times 10^{-5}\ N$

The magnitude of force per unit length of one wire on the other is $2.7945\times 10^{-5}\ N$ and the direction is away from one another

$F_{12}=F_{21}=\frac{\mu_0 i_1i_2}{2\pi r}\\ =\frac{4\pi\times 10^{-7}\times 2.9\times 5.3}{2\pi 0.11}\\ =2.7945\times 10^{-5}\ N$

The magnitude of force per unit length of one wire on the other is $2.7945\times 10^{-5}\ N$ and the direction is towards each other.

7 0

3 years ago

A body travels the first half of the total distance with velocity v and second half with v2 calculate avg velocity

LuckyWell [14K]

Answer:

v = 2 v₁ v₂ / (v₁ + v₂)

Explanation:

The body travels the first half of the distance with velocity v₁. The time it takes is:

t₁ = (d/2) / v₁

t₁ = d / (2v₁)

Similarly, the body travels the second half with velocity v₂, so the time is:

t₂ = (d/2) / v₂

t₂ = d / (2v₂)

The average velocity is the total displacement over total time:

v = d / t

v = d / (t₁ + t₂)

v = d / (d / (2v₁) + d / (2v₂))

v = d / (d/2 (1/v₁ + 1/v₂))

v = 2 / (1/v₁ + 1/v₂)

v = 2 / ((v₁ + v₂) / (v₁ v₂))

v = 2 v₁ v₂ / (v₁ + v₂)

8 0

3 years ago

3 kg

Jet001 [13]

Answer:

The total normal force acting on the system is approximately 58.8 N

Explanation:

The masses arranged in the stack are;

3 kg, 2 kg, and 1 kg

The mass of the stack system, m = 3 kg + 2 kg + 1 kg = 6 kg

Weight = The force of gravity on an object = m·g

Where;

m = The mass of the object

g = The acceleration due to gravity ≈ 9.8 m/s²

∴ The weight of the stack system, W ≈ 6 kg × 9.8 m/s² ≈ 58.8 N

The direction of the weight force = Perpendicular to the surface (acting downwards)

From Newton's third law of motion, the normal force acts perpendicular to the plane and it is equal in magnitude to the force acting perpendicular to the plane

∴ The magnitude of the total normal force acting on the system = The magnitude of the weight of the system ≈ 58.8 N

The (magnitude of the) total normal force acting on the system ≈ 58.8 N

3 0

3 years ago

Holding a metal coat hanger in a campfire and the hanger getting

dedylja [7]

Answer:

That is an example of Radiational transfer

Explanation:

Radiation is a direct transfer of exposer

6 0

3 years ago

Read 2 more answers