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MissTica
1 year ago
7

The integral with respect to time of a force applied to an object is a measure called impulse, and the impulse applied to an obj

ect during a time interval determines its change in momentum during the time interval. The safety of a t-shirt launcher, used to help get crowds cheering at baseball games, is being evaluated. As a first step in the evaluation, engineers consider the design momentum of the launched t-shirts. The springs in the launcher are designed to apply a variable force to a t-shirt over a time interval of t1 = 0.5 s. The force as a function of time is given by F(t) = ať+ b, where a = –28 N/s2 and b = 7.0 N. The momentum of the t-shirt will be its initial momentum (po 0) plus its change in momentum due to the applied impulse: pf = po+SET+ F(t) dt. By applying the given time dependent function for F(t) and performing the integration, which of the following is the correct expression for Pf?
a) tl tl Pf= 0++)16
b) 0+*+*+b)
c) 0+++bt) 0++) ti
Physics
1 answer:
aliina [53]1 year ago
4 0

Answer:

Follows are the solution to this question:

Explanation:

By checking the value in which we have calculated by performing its differentiation of \frac{a}{3}t^3+bt, the correct form of its integer value is calculating  with regard to t, that also provides as expected at^2+b = F(t).

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Every winter I fly home to Chicago. It takes 3 hours. What is my average speed?
Tanya [424]

It depends on where you live when you're not visiting Chicago. We need to know the distance of the trip.

Your average speed on the trip is . . .

(total distance in miles) / (3 hours)

miles per hour

5 0
2 years ago
The drawing shows a hydraulic chamber with a spring (spring constant = 1600 N/m) attached to the input piston and a rock of mass
Triss [41]

Answer:

\Delta x=245\ mm

Explanation:

Given:

  • spring constant of the spring attached to the input piston, k=1600\ N.m^{-1}
  • mass subjected to the output plunger, m=40\ kg

<u>Now, the force due to the mass:</u>

F=m.g

F=40\times 9.8

F=392\ N

<u>Compression in Spring:</u>

\Delta x=\frac{F}{k}

\Delta x=\frac{392}{1600}

\Delta x=0.245\ m

or

\Delta x=245\ mm

8 0
2 years ago
A basketball rolls off a deck that is 3.2 m from the pavement. The basketball lands 0.75 m from the edge of the deck. How fast w
astra-53 [7]

Answer:

d). The value of y should be -32m

Vx=0.92 m/s

Explanation:

Using equation of motion uniform to y motion

x_{f}=x_{o}+v_{o}*t+\frac{1}{2}*a*t^{2}\\x_{o}=0m\\v_{o}=0\\x_{f}=\frac{1}{2}*a*t^{2}\\x_{f}=-3.2m \\a=g=-9.8\frac{m}{s^{2}} \\

So to find t that is the same time for all the motion

t^{2}=\frac{2*x_{f}}{a} \\t=\sqrt{\frac{2*x_{f}}{a}}=\sqrt{\frac{2*-3.2m}{-9.8\frac{m}{s^{2}}}}\\t=0.808s

The value of Xf=-3.2m because the g is negative from the axis

Now in the axis 'x' to find Vx

Vx=\frac{0.75m}{0.808S}\\ Vx=0.92 \frac{m}{s}

5 0
2 years ago
A projectile has an initial horizontal velocity of 15 meters per second and an initial vertical velocity of 25 meters per second
Artyom0805 [142]

Answer:

75 m

Explanation:

The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.

The horizontal component of the velocity of the projectile is

v_x = 15 m/s

and it is constant during the motion;

the total time of flight is

t = 5 s

Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

d= v_x t =(15 m/s)(5 s)=75 m

5 0
2 years ago
Constants Periodic Table Suppose the top surface of the vessel in the figure (Figure 1) is subjected to an external gauge pressu
Gnom [1K]

Answer:

a)  v₁ = √ [2 (P₂-P₀) /ρ + 2 (y₂ -y₁)]

b) Water does not flow,

Explanation:

a) For this exercise we will use Bernoulli's equation, where index 1 is at the exit and index 2 on the surface of the water

            P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

This case does not indicate at the surface pressure is P₂, the pressure at the outlet is P₁ = P₀, the surface velocity is zero v₂ = 0

          P₀ + ½ ρ v₁² + ρ g y₁ = P₂ + 0 + ρ g y₂

           ½ ρ v₁² = P₂-P₀ + ρ (y₂ -y₁)

          v₁² = 2 (P₂-P₀) /ρ + 2 (y₂ -y₁)

          v₁ = √ [2 (P₂-P₀) /ρ + 2 (y₂ -y₁)]

b) Reduce the pressure to SI units

         P₂ = 0.86 atm (1.01 10⁵ Pa / 1 atm) = 0.8686 10⁵ Pa

         P₀ = 1.01 10⁵ Pa

         ρ = 1 10³ kg / m³

Let's calculate

         v₁ = √ [2 (0.8686 -1.01) 10⁵/10³ + 2 (2.6)]

         v₁ = √ [-0.2828 10² + 5.2] = Ra [-23]

Water does not flow, this is because the pressure of the inner part is less than atmospheric, so that the water flows the pressure P₂> = P₀

For example if the pressure P₂ = P₀

         v₁ = √ 5.2

          v₁ = 2.28 m / s

5 0
2 years ago
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