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2 years ago

The distance covered by a car at a time, t is given by x = 20t + 6t4, calculate

Physics

1 answer:

Anni [7]2 years ago

8 0

Answer:

(i) v = 44 m/s

(ii) a = 72 m/s^2

Explanation:

You have the following equation for the potion of a car:

$x=20t+6t^4$

(i) The instantaneous velocity is the derivative of x in time:

$\frac{dx}{dt}=20+(6)(4)t^3=20+24t^3$

for t = 1 is:

$v(t=1)=\frac{dx}{dt}=20+24(1)^3=44m/s$

(ii) The instantaneous acceleration is the derivative of the velocity:

$\frac{dv}{dt}=24(3)t^2=72t^2$

for t = 1

$a(t=1)=\frac{dv}{dt}=72(1)^2=72m/s^2$

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A man attempts to pick up his suitcase of weight w_s by pulling straight up on the handle. (Part A figure) However, he is unable

alexira [117]

Answer:

Part A. The magnitude of the normal force is equal to the magnitude of the weight of the suitcase minus the magnitude of the force of the pull.

Part B. The magnitude of normal force acting on the suitcase is equal to the sum of the weight of the suitcase and the man.

Explanation:

Part A. This is because when the man pulls on the suit upwards, he exerts a force in the upward direction. This takes part of the force of weight of the suitcase and decreases the force the suitcase is exerting on the ground. Thus, the normal force (force exerted by suitcase on the ground) also decreases by the same force as the pull.

Part B. The statements for this part were not given in the question, but the answer reflects what is going to happen in that scenario. Since the man sits on the suitcase, the total weight acting on the ground through the suitcase is that of the suitcase plus the man. Since this force (acting on the ground) is normal force, the statement given in the answer is correct.

8 0

2 years ago

A circular wire loop lies inside a region of space containing a magnetic field. The direction of the magnetic field is out of th

jekas [21]

Answer:

clockwise

Explanation:

According to the law given by Lenz, known as the Lenz law, it is said that a current induced in the circuit which is due to the change in the magnetic field and is so directed so as to oppose the change in the flux and to apply a force in the opposite direction if the force.

Here, as the magnetic field is directed out of the screen, the current flows in the direction which is clockwise in the loop and it opposes the increasing magnetic field.

The clockwise induced current will produce magnetic field in to the screen.

6 0

2 years ago

The weight of an object is the same on two different planets. The mass of planet A is only sixty percent that of planet B. Find

natka813 [3]

Answer:

0.775

Explanation:

The weight of an object on a planet is equal to the gravitational force exerted by the planet on the object:

$F=G\frac{Mm}{R^2}$

where

G is the gravitational constant

M is the mass of the planet

m is the mass of the object

R is the radius of the planet

For planet A, the weight of the object is

$F_A=G\frac{M_Am}{R_A^2}$

For planet B,

$F_B=G\frac{M_Bm}{R_B^2}$

We also know that the weight of the object on the two planets is the same, so

$F_A = F_B$

So we can write

$G\frac{M_Am}{R_A^2} = G\frac{M_Bm}{R_B^2}$

We also know that the mass of planet A is only sixty percent that of planet B, so

$M_A = 0.60 M_B$

Substituting,

$G\frac{0.60 M_Bm}{R_A^2} = G\frac{M_Bm}{R_B^2}$

Now we can elimanate G, MB and m from the equation, and we get

$\frac{0.60}{R_A^2}=\frac{1}{R_B^2}$

So the ratio between the radii of the two planets is

$\frac{R_A}{R_B}=\sqrt{0.60}=0.775$

6 0

2 years ago

What is meant in astronomy by the phrase "adaptive optics?

Sergeu [11.5K]

<h2>Answer: a.The mirrors and eyepiece of a large telescope are spring-loaded to allow them to return quickly to a known position. </h2>

Explanation:

Adaptive optics is a method used in several astronomical observatories to counteract in real time the effects of the Earth's atmosphere on the formation of astronomical images.

This is done through the insertion into the optical path of the telescope of sophisticated deformable mirrors supported by a set of computationally controlled actuators. Thus obtaining clear images despite the effects of atmospheric turbulence that cause the unwanted distortion.

It should be noted that with this technique it is also necessary to have a moderately bright reference star that is very close to the object to be observed and studied. However, it is not always possible to find such stars, so a powerful laser beam is used to point towards the Earth's upper atmosphere and create artificial stars.

7 0

2 years ago

Una cuerda de violin vibra con una frecuencia fundamental de 435 Hz. Cual sera su frecuencia de vibracion si se le somete a una

EleoNora [17]

Answer:

a) f = 615.2 Hz b) f = 307.6 Hz

Explanation:

The speed in a wave on a string is

v = √ T / μ

also the speed a wave must meet the relationship

v = λ f

Let's use these expressions in our problem, for the initial conditions

v = √ T₀ /μ

√ (T₀/ μ) = λ₀ f₀

now it indicates that the tension is doubled

T = 2T₀

√ (T /μ) = λ f

√( 2To /μ) = λ f

√2 √ T₀ /μ = λ f

we substitute

√2 (λ₀ f₀) = λ f

if we suppose that in both cases the string is in the same fundamental harmonic, this means that the wavelength only depends on the length of the string, which does not change

λ₀ = λ

f = f₀ √2

f = 435 √ 2

f = 615.2 Hz

b) The tension is cut in half

T = T₀ / 2

√ (T₀ / 2muy) = f = λ f

√ (T₀ / μ) 1 /√2 = λ f

fo / √2 = f

f = 435 / √2

f = 307.6 Hz

Traslate

La velocidad en una onda en una cuerda es

v = √ T/μ

ademas la velocidad una onda debe cumplir la relación

v= λ f

Usemos estas expresión en nuestro problema, para las condiciones iniciales

v= √ To/μ

√ ( T₀/μ) = λ₀ f₀

ahora nos indica que la tensión se duplica

T = 2T₀

√ ( T/μ) = λf

√ ) 2T₀/μ = λ f

√ 2 √ T₀/μ = λ f

substituimos

√2 ( λ₀ f₀) = λ f

si suponemos que en los dos caso la cuerda este en el mismo armónico fundamental, esto es que la longitud de onda unicamente depende de la longitud de la cuerda, la cual no cambia

λ₀ = λ

f = f₀ √2

f = 435 √2

f = 615,2 Hz

b) La tension se reduce a la mitad

T = T₀/2

RA ( T₀/2μ) = λ f

Ra(T₀/μ) 1/ra 2 = λ f

fo /√ 2 = f

f = 435/√2

f = 307,6 Hz

5 0

2 years ago