Help me please forget about you and she was

A particle travels along the x-axis in such a way that its acceleration at time t is a(t) = t + t2. if it starts at the origin w

Olegator [25]

The acceleration of the particle as a function of time t is
$a(t) = t + t^2$
The velocity of the particle at time t is the integral of the acceleration:
$v(t) = \int {a(t)} \, dt = \frac{t^2}{2} + \frac{t^3}{3} + C$
where the constant C can be found by requiring that the velocity at time t=0 is v=3:
$v(0) = 3$
and we find $C=v_0=3$
so the velocity is
$v(t)=3+ \frac{t^2}{2}+ \frac{t^3}{3}$

The position of the particle at time t is the integral of the velocity:
$x(t)=\int {v(t) } \,dt = 3t + \frac{t^3}{6}+ \frac{t^4}{12} +D$
where D can be found by requiring that the initial position at time t=0 is zero:
x(0)=0
from which we find D=0, so
$x(t)=3t + \frac{t^3}{6}+ \frac{t^4}{12}$

To solve the problem, now we just have to substitute t=5 into x(t) and v(t) to find the position and the velocity of the particle at t=5.

The position is:
$x(5)=3(5) + \frac{5^3}{6}+ \frac{5^4}{12}=87.92$
and the velocity is:
$v(5) = 3+ \frac{5^2}{2}+ \frac{5^3}{3}=57.17$

6 0

3 years ago

Estimate the acceleration due to gravity at the surface of Europa (one of the moons of Jupiter) given that its mass is 4.9×1022k

Taya2010 [7]

Answer: $g=1.97 m/s^2$

Explanation:

Given

mass of Jupiter is $M=4.9\times 10^{22} kg$

Density of Jupiter is same as Earth

$density\ of\ Earth=density\ of\ Jupiter=5510 kg/m^3$

$mass=volume\times density$

considering Jupiter to be sphere of radius r

$M=\frac{4}{3}\times \pi r^3\times \rho$

$r^3=\frac{3M}{\rho \times 4\pi}$

$r^3=\frac{3\times 4.9\times 10^{22}}{5510\times 4\pi}$

$r=(\frac{3\times 4.9\times 10^{22}}{5510\times 4\pi})^{\frac{1}{3}}$

$r=1.28\times 10^6 m$

acceleration due to gravity is given by

$g=\frac{GM}{r^2}$

$g=\frac{6.67\times 10^{-11}\times 4.9\times 10^{22}}{(1.28\times 10^6)^2}$

$g=1.97 m/s^2$

3 0

3 years ago

Why would a ball in outer space move at a constant speed in the same direction?<br> HELp HURRY

Vikki [24]

They air is light and it pushes the ball around

3 0

2 years ago

An American traveler in China carries a transformer to convert China's standard 220 V to 120 V so that she can use some small ap

Arisa [49]

Answer:

(a) The ratio of turns in the primary and secondary coils of her transformer is 1.833

(b) The ratio of input to output current is 0.55

(c) To increase the output voltage, you can either increase the number of turns in the secondary coil (step-up) or increase the input current. Therefore, the Chinese person has to increase the input current of the transformer to achieve an increased output voltage that can power her 220 V appliances.

Explanation:

Given;

input voltage, $V_p$ = 220 V

output voltage, $V_s$ = 120 V

General transformer equation is given as;

$\frac{V_p}{V_s} = \frac{N_p}{N_s} = \frac{I_s}{I_p}$

where;

Np is number of turns in the primary coil

Ns is number of turns in the secondary coil

Is - is the secondary current or output current

Ip - is the primary current or input current

(a) The ratio of turns in the primary and secondary coils of her transformer;

$\frac{N_p}{N_s} = \frac{V_p}{V_s} \\\\\frac{N_p}{N_s} = \frac{220}{120} = 1.833$

(b) The ratio of input to output current;

$\frac{I_p}{I_s} = \frac{V_s}{V_p} \\\\\frac{I_p}{I_s} = \frac{120}{220} \\\\\frac{I_p}{I_s} = 0.55$

(c) To increase the output voltage, you can either increase the number of turns in the secondary coil (step-up) or increase the input current. Therefore, the Chinese person has to increase the input current of the transformer to achieve an increased output voltage that can power her 220 V appliances.

8 0

2 years ago

Compute the density in g/cm? of a piece of metal that has a mass of 0.450 kg and a volume of 52 cm3

Y_Kistochka [10]

Answer:

Ro = 8.65 [g/cm³]

Explanation:

We must remember that density is defined as the ratio of mass to volume.

$Ro=m/V$

where:

m = mass = 0.450 [kg] = 450 [g]

V = volumen = 52 [cm³]

Ro = density [g/cm³]

Now replacing:

$Ro = 450/52\\Ro = 8.65 [g/cm^{3} ]$

8 0

2 years ago