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GarryVolchara [31]
2 years ago
9

The velocity of a an object in linear motion changes from +25 meters per second to +15 meters per second in 2.0 seconds.

Physics
1 answer:
kodGreya [7K]2 years ago
5 0

Answer:

-5.0 m/s^2

Explanation:

Missing question:

What is the object's acceleration?

Solution:

The acceleration of an object is given by

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time taken for the change in velocity

For the object in this problem,

u = +25 m/s

v = +15 m/s

t = 2.0 s

Substituting,

a=\frac{15-25}{2}=-5.0 m/s^2

And the acceleration is negative because its direction is opposite to that of the velocity.

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slava [35]

Answer:

4

Explanation:

Egenuity

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2 years ago
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A mover pushes a 255 kg piano
faust18 [17]

Answer:

0.495 ms^{-2}

Explanation:

According to the newton's second law of motion we can apply F=ma hear

Force = mass * acceleration

(assume the piano is moving left side )

←F = ma

F_(pull)+ F_(push)= M*a\\77.5 + 48.7 = 255 *a\\a = 0.495 ms^{-2}

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2 years ago
Which of the following characterizes Erik Erikson's identity versus role confusion stage?
andrew11 [14]
Identity vs. Role confusion is the 5th stage of Erik Erikson's theory of psychosocial development.

During this stage, adolescents try to explore their independence. They begin to get interested in social interaction and relationships. This is also the time they begin to feel confused and insecure about themselves trying to figure out how to fit in the group. This is the reason why they engage themselves in different roles and activities. 
5 0
2 years ago
A metal sphere with radius R1 has a charge Q1. Take the electric potential to be zero at an infinite distance from the sphere.
Airida [17]

Answer:

Part A :  E =   \frac{1}{4\pi}ε₀ Q₁/R₁² Volt/meter

Part B :  V =  \frac{1}{4\pi}ε₀ Q₁/R₁ Volt

Explanation:

Given that,

Charge distributed on the sphere is Q₁

The radius of sphere is R

₁

The electric potential at infinity is 0

<em>Part A</em>

The space around a charge in which its influence is felt is known in the electric field. The strength at any point inside the electric field is defined by the force experienced by a unit positive charge placed at that point.  

If a unit positive charge is placed at the surface it experiences a force according to the Coulomb law is given by

                          F = \frac{1}{4\pi}ε₀ Q₁/R₁²

Then the electric field at that point is

                                   E =  F/1

                            E =  \frac{1}{4\pi}ε₀ Q₁/R₁²  Volt/meter

Part B

The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against electric forces.

Thus, the electric potential at the surface of the sphere of radius R₁ and charge distribution Q₁ is given by the relation

                           V =  \frac{1}{4\pi}ε₀ Q₁/R₁  Volt

4 0
1 year ago
A pilot in a small plane encounters shifting winds. He flies 26.0 km northeast, then 45.0 km due north. From this point, he flie
cluponka [151]

Answer:

a) v₃ = 19.54 km, b)  70.2º north-west

Explanation:

This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition

vector 1 moves 26 km northeast

let's use trigonometry to find its components

         cos 45 = x₁ / V₁

         sin 45 = y₁ / V₁

         x₁ = v₁ cos 45

         y₁ = v₁ sin 45

         x₁ = 26 cos 45

         y₁ = 26 sin 45

         x₁ = 18.38 km

         y₁ = 18.38 km

Vector 2 moves 45 km north

        y₂ = 45 km

Unknown 3 vector

          x3 =?

          y3 =?

Vector Resulting 70 km north of the starting point

           R_y = 70 km

we make the sum on each axis

X axis

      Rₓ = x₁ + x₃

       x₃ = Rₓ -x₁

       x₃ = 0 - 18.38

       x₃ = -18.38 km

Y Axis

      R_y = y₁ + y₂ + y₃

       y₃ = R_y - y₁ -y₂

       y₃ = 70 -18.38 - 45

       y₃ = 6.62 km

the vector of the third leg of the journey is

         v₃ = (-18.38 i ^ +6.62 j^ ) km

let's use the Pythagorean theorem to find the length

         v₃ = √ (18.38² + 6.62²)

         v₃ = 19.54 km

to find the angle let's use trigonometry

           tan θ = y₃ / x₃

           θ = tan⁻¹ (y₃ / x₃)

           θ = tan⁻¹ (6.62 / (- 18.38))

           θ = -19.8º

with respect to the x axis, if we measure this angle from the positive side of the x axis it is

          θ’= 180 -19.8

          θ’= 160.19º

I mean the address is

          θ’’ = 90-19.8

          θ = 70.2º

70.2º north-west

3 0
2 years ago
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