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2 years ago

What type of foods do bodyguards eat give 5 examples

Physics

1 answer:

Rainbow [258]2 years ago

8 0

They eat rice, chicken, beef, cereal they eat what we eat in a daily basis I guess

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The route followed by a hiker consists of three displacement vectors A with arrow, B with arrow, and C with arrow. Vector A with

kotykmax [81]

Answer:

$D_{B}=1173.98m\\D_{C}=675.29m$

Explanation:

If we express all of the cordinates in their rectangular form we get:

A = (1404.77 , 655.06) m

$B = A + ( -D_{B} *sin(41) , -D_{B} * cos(41) )$

$C = A + B + ( -D_{C} *cos(20) , D_{C} * sin(20) )$

Since we need C to be (0,0) we stablish that:

$C = (0,0) = A + B + ( -D_{C} *cos(20) , D_{C} * sin(20) )$

That way we make an equation system from both X and Y coordinates:

$A_{x} + B_{x} + C_{x} = 0$

$A_{y} + B_{y} + C_{y} = 0$

Replacing values:

$1404.77 - D_{B}*sin(41) - D_{C}*cos(20) = 0$

$655.06 - D_{B}*cos(41) + D_{C}*sin(20) = 0$

With this system we can solve for both Db and Dc and get the answers to the question:

$D_{B}=1173.98m$

$D_{C}=675.29m$

7 0

3 years ago

An inventor develops a stationary cycling device by which an individual, while pedaling, can convert all of the energy expended

Eddi Din [679]

Answer:

Emec = 94050 [J]

Explanation:

In order to solve this problem, we must understand that all thermal energy is converted into mechanical energy.

The thermal energy can be calculated by means of the following expression.

$Q=m*C_{p}*(T_{final}-T_{initial})$

where:

Q = heat [J]

Cp = specific heat of water = 4186 [J/kg*°C]

m = mass = 300 [g] = 0.3 [kg]

T_final = 95 [°C]

T_initial = 20 [°C]

Now we can calculate the heat, replacing the given values:

$Q=0.3*4180*(95-20)\\Q= 94050[J]$

Since all this energy must come from the mechanical energy delivered by the exercise bike, and no energy is lost during the process, the mechanical energy must be equal to the thermal energy.

$Q=E_{mec}\\E_{mec}=94050[J]$

4 0

3 years ago

A ball is tossed from an upper-story window of a building. the ball is given an initial velocity of 8.00 m/s at an angle of 20.0

otez555 [7]

A) The motion of the ball consists of two indipendent motions on the horizontal (x) and vertical (y) axis. The laws of motion in the two directions are:
$x(t)=v_0 \cos \alpha t$
$y(t)=h-v_0 \sin \alpha t - \frac{1}{2}gt^2$
where
- the horizontal motion is a uniform motion, with constant speed $v_0 \cos \alpha$ , where $v_0 = 8.00 m/s$ and $\alpha=20.0^{\circ}$
- the vertical motion is an uniformly accelerated motion, with constant acceleration $g=9.81 m/s^2$ , initial position h (the height of the building) and initial vertical velocity $v_0 \sin \alpha$ (with a negative sign, since it points downwards)

The ball strikes the ground after a time t=3.00 s, so we can find the distance covered horizontally by the ball by substituting t=3.00 s into the equation of x(t):
$x(3.00 s)=v_0 \cos \alpha t=(8 m/s)(\cos 20^{\circ})(3.0 s)=22.6 m$

b) To find the height from which the ball was thrown, h, we must substitute t=3.00 s into the equation of y(t), and requiring that y(3.00 s)=0 (in fact, after 3 seconds the ball reaches the ground, so its vertical position y(t) is zero). Therefore, we have:
$0=h-v_0 \sin \alpha t - \frac{1}{2}gt^2$
which becomes
$h=(8 m/s)(\sin 20^{\circ})(3.0 s)+ \frac{1}{2}(9.81 m/s^2)(3.0 s)^2=52.3 m$

c) We want the ball to reach a point 10.0 meters below the level of launching, so we want to find the time t such that
$y(t)=h-10$
If we substitute this into the equation of y(t), we have
$h-10 = h-v_0 \sin \alpha t- \frac{1}{2}gt^2$
$\frac{1}{2}gt^2+v_0 \sin \alpha t -10 =0$
$4.9 t^2 +2.74 t-10 =0$
whose solution is $t=1.18 s$ (the other solution is negative, so it has no physical meaning). Therefore, the ball reaches a point 10 meters below the level of launching after 1.18 s.

4 0

3 years ago

Running at 2 m/s, Bruce the 45 kg quarterback, collides with Biff, the 90 kg tackle, who is traveling at 7 m/s in the other dire

melomori [17]

Given data

*The mass of Bruce is m_1 = 45 kg

*The initial velocity of the Bruce is u_1 = 2 m/s

*The mass of the biff is m_2 = 90 kg

*The initial velocity of the Biff is u_2 = -7 m/s

*The final velocity of the first glider is v_2 = -1 m/s

According to the law of conservation of linear momentum, the total linear momentum of a system remains constant

Applying the law of conservation of momentum as

$\begin{gathered} p_i=p_f \\ m_1u_1+m_2u_2=m_1v_1+m_2v_2 \\ v_1=\frac{m_1u_1+m_2u_2-m_2v_2_{}_{}_{}_{}}{m_1} \end{gathered}$

Substitute the known values in the above expression as

$\begin{gathered} v_1=\frac{(45)(2)+(90)(-7)-(90)(-1)}{45} \\ =-10\text{ m/s} \end{gathered}$

Hence, the speed of the bruce knock backwards is v_1 = -10 m/s

3 0

1 year ago

SUPPOSE THAT YOURE FACING A STRAIGHT CURRENT CARRYING CONDUCTOR, AND the current isfowing towards you. The lines of magnetic for

enot [183]

According to the right hand grip rule; if you point your right thumb in the direction of the current flow and curl your fingers around the current carrying conductor, the fingers will point in the direction of the circular magnetic field around the conductor. Therefore,if the the current carrying carrying conductor is held with the current flowing towards you, using right hand grip rule the lines of magnetic field will act towards a counterclockwise direction or anticlockwise direction.

7 0

3 years ago