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4 years ago

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A 10 kg mass rests on a table. What acceleration will be generated when a force of 20 N is applied and encounters a frictional f

orce of 15 N? A) 0.5 m/s2 B) 3.5 m/s2 C) 5 m/s2 D) 35 m/s2

1 answer:

kolbaska11 [484]4 years ago

3 0

Yeah. I think that its D too

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Pls answer this question from physics

KatRina [158]

1. Anil leaves home at 0905

2. Anil stays at the park for 10 mins.

3. The distance of Anil's home from the park is 7 km

4. Anil's speed is 14 km/h

The question is a distance-time graph

<h3>What is a distance-time graph?</h3>

A distance-time graph show the relationship between distanced moved by an object and time.

<h3>1. When did Anil leave home?</h3>

Anil left home at the time when the distance is zero. That is, the time at which the graph cuts the time axis. That time is 0905.

So, Anil leaves home at 0905

<h3>2. How many minutes did Anil wait at the park?</h3>

Anil waited at the park when her distance is constant. That is the part of the graph that is parallel to the time axis.

It starts at 0935 and ends at 0945. So, the time anil stays at the park is t = 0945 - 0935 = 10 min.

So, Anil stays at the park for 10 mins.

<h3>3. What is the distance of Anil's home to the park?</h3>

The distance from Anil's home to the park is the maxiimum distance she moves until she reaches the park. From the graph, this is 7 km.

So, the distance of Anil's home from the park is 7 km

<h3>4. What was Anil's speed in his journey to the park?</h3>

Anil's speed in his journey to the park is the gradient of the first slopy line between the points (0905, 0) and (0935, 7)

The gradient, m = (d' - d)/(t' - t) where

(d,t) = (0905, 0) and
(t', d') = (0935, 7)

So, m = (d' - d)/(t' - t)

= (7 km - 0 km)/(0935 - 0905) min

= 7 km/30 min

= 7 km/30 min × 60 min/1 hr

= 7 km × 2/hr

= 14 km/h

So, Anil's speed is 14 km/h

Learn more about distance-time graph here:

brainly.com/question/27956440

#SPJ1

6 0

2 years ago

What are Kepler’s Laws of Planetary Motion? How can each one of them be applied to the 8 planets of the Solar System, using exam

faust18 [17]

Answer:

Explanation:

As per the Kepler's law of planetary motions :

1- The orbits are of elliptical shape having two foci and Sun is located on one foci.

2- The line segment that joins a planet and the Sun sweeps out equal are at equal interval of time.

3- The orbital period square is directly proportional to the cube of semi major axis of its orbit.

Kepler's law are applied on each of the planets of our solar system as the distance of the Sun from the planet is calculated through this.

For example : from Kepler's first law we can see the eccentricity of the Earth's orbit is 0.0167.

6 0

3 years ago

Which two characteristics make electromagnetic waves effective in communication?

Talja [164]

Answer:

A. They do not require a medium to carry information.

D. They carry information very quickly.

Explanation:

The two characteristics that makes electromagnetic waves effective in communication is that they do not require a medium to carry information and they move information very quickly.

Electromagnetic waves are produced by the vibration between electrical and magnetic fields.
These waves are notable for being able to propagate without requiring any medium for their movement.
They also have very fast speed.
The speed of all electromagnetic waves is equal to that of light.

3 0

3 years ago

Collisions between atoms are often elastic, but sometimes inelastic collisions occur, and the lost kinetic energy can become int

Galina-37 [17]

Answer:

a) E = 6.4 1019 J b) v = 0.69 10⁴4 m / s

Explication

a) convert E = 4.0 eV

1 eV = 1.6 10⁻¹⁹ J

E = 4.0 eV (1.6 10⁻¹⁹ J / 1 eV)

E = 6.4 10⁻¹⁹ J

b) Suppose we have a frontal shock and all the kinetic energy of oxygen is transferred to Cs

Ei = K = ½ m v²

Ef = 6.4 10⁻¹⁹ J

½ m v² = 6.4 10⁻¹⁹

The oxygen mass of the periodic table is

PA = 15,999 u

1u = 1.660 10⁻²⁷ kg

Pa = 15,999 1,660 10⁻²⁷ kg

m= Pa = 26,558 10⁻²⁷ kg

Let's calculate the speed

v2 = 2 / m 6.4 10⁻¹⁹

v2 = 2 / 26,558 10⁻²⁷ 6.4 10⁻¹⁹ =

v = √0.4819 10⁸

v = 0.69 10⁴4 m / s

4 0

4 years ago

Light of wavelength 656 nm and 410 nm emitted from a hot gas of hydrogen atoms strikes a grating with 5300 lines per centimeter.

padilas [110]

Answer:

$20.32^{\circ}$ and $44.08^{\circ}$

$12.56^{\circ}$ and $25.77^{\circ}$

Explanation:

$\lambda$ = Wavelength

$\theta$ = Angle

m = Order

Distance between grating is given by

$d=\dfrac{1}{5300}\\\Rightarrow d=0.0001886\ \text{cm}$

$\lambda=656\ \text{nm}$

We have the relation

$d\sin\theta=m\lambda\\\Rightarrow \theta=\sin^{-1}\dfrac{m\lambda}{d}$

m = 1

$\theta=\sin^{-1}\dfrac{1\times 656\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=20.35^{\circ}$

m = 2

$\theta=\sin^{-1}\dfrac{2\times 656\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=44.08^{\circ}$

The first and second order angular deflection is $20.32^{\circ}$ and $44.08^{\circ}$

$\lambda=410\ \text{nm}$

m = 1

$\theta=\sin^{-1}\dfrac{1\times 410\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=12.56^{\circ}$

m = 2

$\theta=\sin^{-1}\dfrac{2\times 410\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=25.77^{\circ}$

The first and second order angular deflection is $12.56^{\circ}$ and $25.77^{\circ}$ .

4 0

3 years ago