The gravitational pull between two objects depends on their mass and distance. What is meant by distance?

which of the following values can you find by calculating the slope of a graph of velocity versus time position . displacement .

statuscvo [17]

The correct answer is: acceleration.

In fact, the slope of the graph of velocity versus time corresponds to the ratio between the variation of v and the variation of t:
$\frac{\Delta v}{\Delta t}$
But this is exactly the definition of acceleration: $a= \frac{\Delta v}{\Delta t}$ , so acceleration is the correct answer.

5 0

3 years ago

Compare the de Broglie wavelength of an electron moving at 1.30×107 miles per hour (5.81×106 m/s) to that of a bullet moving at

lesya [120]

Answer:

electron λ = 12.5 nm , bullet λ = 1.11 10⁻³³ m and golf ball λ = 4.7 10⁻³⁴ m

Explanation:

The Broglie wave duality principle states that all matter has wave and particle properties, it is expressed by the equation

p = h / λ

Where lam is called broglie wavelength

Let's use the definition of momentum

p = mv

Let's calculate the wavelengths

-Electron

mv = h /λ

λ = h / mv

λ = 6.63 10⁻³⁴ / (9.1 10⁻³¹ 5.81 10⁶)

λ = 1.25 10⁻¹⁰ m

λ = 12.5 nm

This is the X-ray region

-bullet

λ = 6.63 10⁻³⁴ / (1.90 10⁻³ 313)

λ = 1.11 10⁻³³ m

It is too small, only particle characteristics are observed

-Golf ball

λ = 6.63 10⁻³⁴ / (4.50 10⁻² 31.3)

λ = 4.7 10⁻³⁴ m

Too small, only particle characteristics are visible

7 0

3 years ago

As a train accelerates away from a station, it reaches a speed of 4.6 m/s in 5.2 s. If the train's acceleration remains constant

Svet_ta [14]

Answer:

Vf = 10.76 m/s

Explanation:

Train kinematics

The train moves with uniformly accelerated movement

$V_f = V_o + a*t$ Formula (1)

Vf: Final speed (m/s)

V₀: Inital speed (m/s)

t: time in seconds (s)

a: acceleration (m/s²)

Movement from t = 0 to t = 5.2s

We replace in formula (1)

4.6 = 0 + a*5.2

a = 4.6/5.2 = 0.88 m/s²

Movement from t = 5.2s to t = 5.2s + 7s = 12.2s

We replace in formula (1)

$V_f = 4.6 + 0.88*7$

Vf = 10.76 m/s

3 0

3 years ago

A scientist uses a camera to study the stars.

likoan [24]

Answer:

All of teh above except A

Explanation:

3 0

3 years ago

A 100-turn, 2.0-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 60∘ away from vertical increases fro

bixtya [17]

Answer:

The induced emf is $|\epsilon|=0.0261 V$

Explanation:

From the question we are told that

The number of turn is $N = 100$

The diameter of the coil is $d = 2.0 cm = 2.0 *10^{-2} m$

The uniform magnetic at initial is $B_i = 0.50 T$

The uniform magnetic at initial is $B_f = 1.50 T$

The time taken is $t = 0.60s$

The angle the magnetic field makes with vertical is $\theta = 60^o$

Generally induced emf is mathematically represented as

$\epsilon = -N \frac{d \o}{dt}$

where $d \o$ is the change magnetic flux

Magnetic flux is mathematically represented as

$\O = \= B \cdot \= A$

$= BA cos \theta$

Substituting this above

$\epsilon = -N \frac{d (BA cos \theta)}{dt}$

$\epsilon = -N A \frac{d (B cos \theta)}{dt}$

Where B is the magnetic field and A is the area which is mathematically evaluated as

$A = \frac{\pi d^2}{4}$

Substituting values

$A = \frac{\pi (2.0 *10^{-2})^2}{4}$

$A= 3.142*10^{-4}m^2$

From the equation of emf

$\epsilon = -N A \frac{d (B cos \theta)}{dt}$

dB = $B_2 -B_1$

So

$|\epsilon| = N A \frac{ (B_2 -B_1 cos \theta)}{dt}$

substituting values

$|\epsilon| = 100(3.142*10^{-4}) \frac{ (1.50 -0.50 cos(60))}{0.60}$

$|\epsilon|=0.0261 V$

6 0

3 years ago