All categories

2 years ago

Acceleration is defined as the rate of change of position true or false

Physics

2 answers:

Svetradugi [14.3K]2 years ago

7 0

Answer:

True

Explanation:

Likurg_2 [28]2 years ago

6 0

The answer is : True

You might be interested in

When landing from a jump, a basketball player of mass 82 kg has a velocity of 1.2 m/s right before they hit the ground. The play

saw5 [17]

Answer:

2361.6N

Explanation:

Mass of player = 82kg

Velocity = 1.2m/s

Kinetic energy of player:

= 1/2mv²

= 1/2*82*1.2²

= 41x1.44

= 59.04J

Final kinetic energy = 0

Change in kinetic energy

|∆k| = |0-59.04|

= 59.04

Workdone by the feet = fd

d = 0.025

Fd = 59.04

F = 59.04/0.025

= 2361.6N

This is his average force.

6 0

2 years ago

Which statement below concerning the photoelectric effect is true? View Available Hint(s) Which statement below concerning the p

kakasveta [241]

There is a threshold frequency for each metal, and only light of a frequency higher than this threshold causes electrons to be emitted from the metal surface.

8 0

3 years ago

Suppose that the velocity (in meters per second) of a sky diver falling near the Earth's surface is given by the following expon

Mice21 [21]

Answer:

4.7 s

Explanation:

The complete question is presented in the attached image to this solution.

v(t) = 61 - 61e⁻⁰•²⁶ᵗ

At what time will v(t) = 43 m/s?

We just substitute 43 m/s into the equation for the velocity of the diver and solve for t.

43 = 61 - 61e⁻⁰•²⁶ᵗ

- 61e⁻⁰•²⁶ᵗ = 43 - 61 = -18

e⁻⁰•²⁶ᵗ = (18/61) = 0.2951

In e⁻⁰•²⁶ᵗ = In 0.2951 = -1.2205

-0.26t = -1.2205

t = (1.2205/0.26) = 4.694 s = 4.7 s to the nearest tenth.

Hope this Helps!!!

6 0

2 years ago

A 22.0 kg bucket of concrete is connected over a very light frictionless pulley to a 375 N box on the roof of a building as show

Veronika [31]

Answer:

vf = 3.27[m/s]

Explanation:

In order to solve this problem we must analyze each body individually and find the respective equations. The free body diagram of each body (box and bucket) should be made, in the attached image we can see the free body diagrams and the respective equations.

With the first free body diagram, we determine that the tension T should be equal to the product of the mass of the box by the acceleration of this.

With the second free body diagram we determine another equation that relates the tension to the acceleration of the bucket and the mass of the bucket.

Then we equalize the two stress equations and we can clear the acceleration.

a = 3.58 [m/s^2]

As we know that the bucket descends 1.5 [m], this same distance is traveled by the box, as they are connected by the same rope.

$x = \frac{1}{2} *a*t^{2}\\1.5 = \frac{1}{2}*(3.58) *t^{2} \\t = 0.91 [s]$

And the speed can be calculated as follows:

$v_{f}=v_{o}+a*t\\v_{f}=0+(3.58*0.915)\\v_{f}= 3.27[m/s]$

7 0

2 years ago

Q.3. The equivalent resistance across AB is:<br> (a)1<br> (c)2<br> (b)3<br> (d)4

sp2606 [1]

Answer:

1 ohm

Explanation:

First of all, the equivalent resistance for two resistors (r₁ and r₂) in parallel is given by:

1 / Eq = (1 / r₁) + (1 / r₂)

The equivalent resistance for resistance for two resistors (r₁ and r₂) in series is given by:

Eq = r₁ + r₂

Hence as we can see from the circuit diagram, 2Ω // 2Ω, and 2Ω // 2Ω, hence:

1/E₁ = 1/2 + 1/2

1/E₁ = 1

E₁ = 1Ω

1/E₂ = 1/2 + 1/2

1/E₂ = 1

E₂ = 1Ω

This then leads to E₁ being in series with E₂, hence the equivalent resistance (E₃) of E₁ and E₂ is:

E₃ = E₁ + E₂ = 1 + 1 = 2Ω

The equivalent resistance (Eq) across AB is the parallel combination of E₃ and the 2Ω resistor, therefore:

1/Eq = 1/E₃ + 1/2

1/Eq = 1/2 + 1/2

1/Eq = 1

Eq = 1Ω

7 0

2 years ago