2 general categories of factors that determine how we look and behave, and how do they influence us?

At its lowest setting a centrifuge rotates with an angular speed of calculate the angular velocity

I am Lyosha [343]

No answer is possible until we know the number that belongs after the words "... angular speed of ".

8 0

3 years ago

A 51.0 kg cheetah accelerates from rest to its top speed of 31.7 m/s. HINT (a) How much net work (in J) is required for the chee

andrey2020 [161]

(a) $2.56\cdot 10^4 J$

The work-energy theorem states that the work done on the cheetah is equal to its change in kinetic energy:

$W= \Delta K = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

where

m = 51.0 kg is the mass of the cheetah

u = 0 is the initial speed of the cheetah (zero because it starts from rest)

u = 31.7 m/s is the final speed

Substituting, we find

$W=\frac{1}{2}(51.0 kg)(31.7 m/s)^2 - \frac{1}{2}(51.0 kg)(0)^2=2.56\cdot 10^4 J$

(b) 6.1 cal

The conversion between calories and Joules is

1 cal = 4186 J

Here the energy the cheetah needs is

$E=2.56\cdot 10^4 J$

Therefore we can set up a simple proportion

$1 cal : 4186 J = x : 2.56\cdot 10^4 J$

to find the equivalent energy in calories:

$x=\frac{(1 cal)(2.56\cdot 10^4 J)}{4186 J}=6.1 cal$

3 0

3 years ago

From the window of a building, a ball is tossed from a height y0 above the ground with an initial velocity of 8.10 m/s and angle

GarryVolchara [31]

Answer:

Part a)

$y = 88.5 m$

Part b)

$v_x = 7.7 m/s$

$v_y = 2.5 m/s$

Part c)

equation of position in x direction is given as

$x = v_x t$

equation of position in y direction is given as

$y = v_y t + \frac{1}{2}gt^2$

Part d)

$x = 30.8 m$

Part e)

H = 88.5 m

Part f)

t = 1.2 s

Explanation:

As we know that ball is projected with speed

v = 8.10 m/s at an angle 18 degree below the horizontal

so we will have

$v_x = 8.10 cos18$

$v_x = 7.7 m/s$

$v_y = 8.10 sin18$

$v_y = 2.5 m/s$

Part a)

Since it took t =4 s to reach the ground

so its initial y coordinate is given as

$y = v_y t + \frac{1}{2}a_y t^2$

$y = 2.5(4) + \frac{1}{2}(9.81)(4^2)$

$y = 88.5 m$

Part b)

components of the velocity is given as

$v_x = 8.10 cos18$

$v_x = 7.7 m/s$

$v_y = 8.10 sin18$

$v_y = 2.5 m/s$

Part c)

equation of position in x direction is given as

$x = v_x t$

equation of position in y direction is given as

$y = v_y t + \frac{1}{2}gt^2$

Part d)

distance where it will strike the floor is given as

$x = v_x t$

$x = 7.7 \times 4$

$x = 30.8 m$

Part e)

Height from which it is thrown is same as initial y coordinate of the ball

so it is given as

H = 88.5 m

Part f)

time taken by ball to reach 10 m below is given as

$y = v_y t + \frac{1}{2}gt^2$

$10 = 2.5t + \frac{1}{2}(9.81) t^2$

t = 1.2 s

7 0

3 years ago

Help me... my science is horrible. I will mark brainliest, would prefer fast answer...

Elena L [17]

vacuum walls: prevents heat loss by comduction and radiation

silvered walls prevent loss by radiation

ceramic base prevents loss by conduction

cap prevents by radiation and convection and reduces by comduction

7 0

3 years ago

Two girls,(masses m1 and m2) are on roller skates and stand at rest, close to each other and face-to-face. Girl 1 pushes squarel

Arturiano [62]

Answer:

$\vec{v}_1 = -\frac{\vec{v}_2m_2}{m_1}$

Explanation:

The center of mass of the system (two girls) is constant, as the velocity of the center of mass of the system is also constant.

$\vec{v}_{cm} = \frac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2}$

<u>The initial velocity of the system is zero, since both girls are at rest.</u> So the velocity of the total system at any point should be zero as well.

$0 = \frac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2}\\\vec{v}_1 = -\frac{\vec{v}_2m_2}{m_1}$

This is true, because there is no friction between the girls and the ground. Otherwise, the velocity of the center of mass wouldn't be constant.

5 0

3 years ago