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aleksandrvk [35]
3 years ago
11

A water balloon is thrown at an angle of 30° with a velocity of 40 m/s. How long do you have to move out of the way if the ballo

on was 20m away from you when it was thrown?
Physics
1 answer:
boyakko [2]3 years ago
6 0

Answer:

t=0.58s

Explanation:

We can consider only the horizontal component since the horizontal component of the velocity is constant (v_x=vcos\theta), and we want to know how much time it takes for the balloon to travel a horizontal distance d_x at that speed. The definition of (horizontal) velocity is v_x=\frac{d_x}{t}, so we have:

t=\frac{d_x}{v_x}=\frac{d_x}{vcos\theta}=\frac{(20m)}{(40m/s)cos(30^{\circ})}=0.58s

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Answer:

-mg = -7.52*9.8 = - 73.696N

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Suppose you walk into a sauna that has an ambient temperature of 48.0°C. (a) Calculate the rate of heat transfer in watts to you
GenaCL600 [577]

Answer:99.84 W

Explanation:

Given

Ambient temperature T_{\infty }=48^{\circ}C\approx 321\ K

Temperature of skin T=37^{\circ}C\approx 310\ K

Skin emissivity \epsilon =0.98

Surface area A=1.30\ m^2

Rate of heat transfer is Q=\epsilon A\sigma (T_0^4-T^4)

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Q=99.84\ W

4 0
3 years ago
Suppose you have a 0.750-kg object on a horizontal surface connected to a spring that has a force constant of 150 N/m. There is
marshall27 [118]

Answer:

x=0.0049\ m= 4.9\ mm

d=0.01153\ m=11.53\ mm

Explanation:

Given:

  • mass of the object, m=0.75\ kg
  • elastic constant of the connected spring, k=150\ N.m^{-1}
  • coefficient of static friction between the object and the surface, \mu_s=0.1

(a)

Let x be the maximum distance of stretch without moving the mass.

<em>The spring can be stretched up to the limiting frictional force 'f' till the body is stationary.</em>

f=k.x

\mu_s.N=k.x

where:

N = m.g = the normal reaction force acting on the body under steady state.

0.1\times (9.8\times 0.75)=150\times x

x=0.0049\ m= 4.9\ mm

(b)

Now, according to the question:

  • Amplitude of oscillation, A= 0.0098\ m
  • coefficient of kinetic friction between the object and the surface, \mu_k=0.085

Let d be the total distance the object travels before stopping.

<em>Now, the energy stored in the spring due to vibration of amplitude:</em>

U=\frac{1}{2} k.A^2

<u><em>This energy will be equal to the work done by the kinetic friction to stop it.</em></u>

U=F_k.d

\frac{1}{2} k.A^2=\mu_k.N.d

0.5\times 150\times 0.0098^2=0.0850 \times 0.75\times 9.8\times d

d=0.01153\ m=11.53\ mm

<em>is the total distance does it travel before stopping.</em>

7 0
3 years ago
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