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2 years ago

If the loss of 3500 kcal is equal to a loss of 1.0 lb, how many days will it take charles to lose 5.0 lb

1 answer:

4 0

We are missing an important piece of information needed to answer this question: the number of kcal Charles losses per day. However, we can come up with a general equation in which kcal/day is the only independent variable.

We know that it takes 3500 kcal to lose one pound. To lose 5 pounds, Charles needs to lose 5 x 3500 kcal = 17,500 kcal.

To find how many days it takes Charles to lose 17,500 kcal (5 pounds), we must divide that amount by the number of kcal Charles loses per day.
Here is the equation to calculate that number

Number of days= 17500 / (kcal per day)

If given calories, remember that 1000 calories = 1 kcal, and .001 kcal = 1 cal

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what is the acceleration of a bowling ball that starts at rest and moves 300m down the gutter in 22.4 sec

exis [7]

<span>Acceleration is the change in velocity divided by time taken. It has both magnitude and direction. In this problem, the change in velocity would first have to be calculated. Velocity is distance divided by time. Therefore, the velocity here would be 300 m divided by 22.4 seconds. This gives a velocity of 13.3928 m/s. Since acceleration is velocity divided by time, it would be 13.3928 divided by 22.4, giving a final solution of 0.598 m/s^2.</span>

7 0

2 years ago

Un niño de 25 kg corre por un jardín con una velocidad de 2.5 m/s de forma que su trayectoria es tangente al borde de un carruse

Schach [20]

Answer:

La velocidad angular del niño y del carrusel cuando se mueven juntos es 0.208 radianes por segundo.

Explanation:

Asumamos que tanto el niño como el carrusel no tienen carga externa aplicada sobre aquellos, de modo que se puede aplicar el Principio de Conservación de la Cantidad de Movimiento Angular:

$m\cdot v \cdot R = (m\cdot R^{2}+I)\cdot \omega$ (1)

Donde:

$m$ - Masa del niño, medida en kilogramos.

$v$ - Velocidad lineal inicial del niño, medida en metros por segundo.

$R$ - Radio máximo del carrusel, medida en metros.

$I$ - Momento de inercia del carrusel, medida en kilogramo-metros cuadrados.

$\omega$ - Velocidad angular final del sistema niño-carrusel, medida en radianes por segundo.

Si sabemos que $m = 25\,kg$ , $v = 2.5\,\frac{m}{s}$ , $R = 2\,m$ y $I = 500\,kg\cdot m^{2}$ , tenemos que la velocidad angular final es:

$\omega = \frac{m\cdot v\cdot R}{m\cdot R^{2}+I}$

$\omega = \frac{(25\,kg)\cdot \left(2.5\,\frac{m}{s} \right)\cdot (2\,m)}{(25\,kg)\cdot (2\,m)^{2}+500\,kg\cdot m^{2}}$

$\omega = 0.208\,\frac{rad}{s}$

La velocidad angular del niño y del carrusel cuando se mueven juntos es 0.208 radianes por segundo.

4 0

2 years ago

A 2530-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that

gavmur [86]

Answer:

A = 1.4 m/s²

B = -0.10493 m/s³

a = 1.29507 m/s²

T = 28095.8271 N

T = 1.13198 W

Explanation:

t = Time taken

g = Acceleration due to gravity = 9.81 m/s²

The equation

$v(t)=At+Bt^2$

Differentiating with respect to time

$\frac{dv}{dt}=\frac{d(At+Bt^2)}{dt}\\\Rightarrow 1.4=A+2Bt$

At t = 0

$1.4=A$

Hence, A = 1.4 m/s²

$B=\frac{v-At}{t^2}\\\Rightarrow B=\frac{2.18-1.4\times 1.8}{1.8^2}\\\Rightarrow B=-0.10493\ m/s^3$

B = -0.10493 m/s³

At t = 5 seconds

$a=1.4+2\times -0.010493\times 5=1.29507\ m/s^2$

a = 1.29507 m/s²

$T=m(a+g)\\\Rightarrow T=2530(1.29507+9.81)\\\Rightarrow T=28095.8271\ N$

T = 28095.8271 N

Weight of rocket

$W=2530\times 9.81=24819.9\ N$

$\frac{T}{W}=\frac{28095.8271}{24819.9}\\\Rightarrow \frac{T}{W}=1.13198\\\Rightarrow T=1.13198W$

T = 1.13198 W

3 0

2 years ago

A turntable that is initially at rest is set in motion with a constant angular acceleration α. What is the magnitude of the angu

bekas [8.4K]

Explanation:

If the turntable starts from rest and is set in motion with a constant angular acceleration α. Let $\omega$ is the angular velocity of the turntable. We know that the rate of change of angular velocity is called the angular acceleration of an object. Its formula is given by :

$\alpha =\dfrac{\omega_f-\omega_i}{t}$

$\alpha =\dfrac{\omega-0}{t}$

$\alpha =\dfrac{\omega}{t}$

$t=\dfrac{\omega}{\alpha }$ ............(1)

Using second equation of kinematics as :

$\theta=\omega_i t+\dfrac{1}{2}\alpha t^2$

$\theta=\dfrac{1}{2}\alpha t^2$

Using equation (1) in above equation

$\theta=\dfrac{1}{2}\times \dfrac{\omega^2}{\alpha }$

In one revolution, $\theta=4\pi$ (in 2 revolutions)

$4\pi =\dfrac{1}{2}\times \dfrac{\omega^2}{\alpha }$

$\omega=\sqrt{8\pi \alpha}$

$\omega=2\sqrt{2\pi \alpha}$

Hence, this is the required solution.

6 0

2 years ago

Read 2 more answers

What is the speed of a beam of electrons when the simultaneous influence of an electric field of 1.56×104v/m and a magnetic fiel

sashaice [31]

1) $3.38\cdot 10^6 m/s$

When both the electric field and the magnetic field are acting on the electron normal to the beam and normal to each other, the electric force and the magnetic force on the electron have opposite directions: in order to produce no deflection on the electron beam, the two forces must be equal in magnitude

$F_E = F_B\\qE = qvB$

where

q is the electron charge

E is the magnitude of the electric field

v is the electron speed

B is the magnitude of the magnetic field

Solving the formula for v, we find

$v=\frac{E}{B}=\frac{1.56\cdot 10^4 V/m}{4.62\cdot 10^{-3} T}=3.38\cdot 10^6 m/s$

2) 4.1 mm

When the electric field is removed, only the magnetic force acts on the electron, providing the centripetal force that keeps the electron in a circular path:

$qvB=m\frac{v^2}{r}$

where m is the mass of the electron and r is the radius of the trajectory. Solving the formula for r, we find

$r=\frac{mv}{qB}=\frac{(9.1 \cdot 10^{-31} kg)(3.38\cdot 10^6 m/s)}{(1.6\cdot 10^{-19} C)(4.62\cdot 10^{-3}T)}=4.2\cdot 10^{-3} m=4.1 mm$

3) $7.6\cdot 10^{-9}s$

The speed of the electron in the circular trajectory is equal to the ratio between the circumference of the orbit, $2 \pi r$ , and the period, T:

$v=\frac{2\pi r}{T}$

Solving the equation for T and using the results found in 1) and 2), we find the period of the orbit:

$T=\frac{2\pi r}{v}=\frac{2\pi (4.1\cdot 10^{-3} m)}{3.38\cdot 10^6 m/s}=7.6\cdot 10^{-9}s$

7 0

2 years ago