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MissTica
2 years 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]2 years 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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At a given point on a horizontal streamline in flowing air, the static pressure is â2.0 psi (i.e., a vacuum) and the velocity is
Nastasia [14]
At a point on the streamline, Bernoulli's equation is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = density of air, 0.075 lb/ft³ (standard conditions)
g = 32 ft/s²

Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s

Point 2 (stagnation):
At the stagnation point, the velocity is zero.

The density remains constant.
Let p₂ = pressure at the stagnation point.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
     = 314.37 lb/ft²
     = 314.37/144 = 2.18 lb/in²

Answer: 2.2 psi

5 0
2 years ago
Suppose two astronauts on a spacewalk are floating motionless in space, 3.0 m apart. Astronaut B tosses a 15.0 kg IMAX camera to
marta [7]

Answer:

\frac{ 112.5}{15+m_{A}}=v_{f}

(we need the mass of the astronaut A)

Explanation:

We can solve this by using the conservation law of the linear momentum P. First we need to represent every mass as a particle. Also we can simplify this system of particles by considering only the astronaut A with an initial speed v_{iA} of 0 m/s and a mass m_{A} and the IMAX camera with an initial speed v_{ic} of 7.5 m/s and a mass m_{c} of 15.0 kg.

The law of conservation says that the linear momentum P (the sum of the products between all masses and its speeds) is constant in time. The equation for this is:

P_{i}=p_{ic}+p_{iA}\\P_{i}=m_{c}v_{ic}+m_{A} v_{iA}\\P_{i}=15*7.5 + m_{A}*0\\P_{i}=112.5 \frac{kg.m}{s}

By the law of conservation we know that P_{i} =P_{f}

For P_{f} (final linear momentum) we need to treat the collision as a plastic one (the two particles stick together after the encounter).

So:

P_{i} =P_{f}=112.5\\

112.5=(m_{c}+m_{A})v_{f}\\\frac{ 112.5}{m_{c}+m_{A}}=v_{f}\\\frac{ 112.5}{15+m_{A}}=v_{f}

3 0
2 years ago
Consider a space shuttle which has a mass of about 1.0 x 105 kg and circles the Earth at an altitude of about 200.0 km. Calculat
kodGreya [7K]

Answer:

1.6675×10^-16N

Explanation:

The force of gravity that the space shuttle experiences is expressed as;

g = GM/r²

G is the gravitational constant

M is the mass = 1.0 x 10^5 kg

r is the altitude = 200km = 200,000m

Substitute into the formula

g = 6.67×10^-11 × 1.0×10^5/(2×10^5)²

g = 6.67×10^-6/4×10^10

g = 1.6675×10^{-6-10}

g = 1.6675×10^-16N

Hence the force of gravity experienced by the shuttle is 1.6675×10^-16N

7 0
2 years ago
The acceleration due to gravity at the north pole of Neptune is approximately 11.2 m/s2. Neptune has mass 1.02×1026 kg and radiu
larisa [96]

Answer: (a) The gravitational force on the object at the North Pole of Neptune is 51.7N

(b) The apparent weight of the object at Neptune's equator is 50.4N

Explanation: Please see the attachments below

5 0
2 years ago
Alex goes cruising on his dirt bike. He rides 700m north, 300m east, 400 m north, 600m west, 1200m south, 300m east and finally
Bess [88]

Find Displacement and Distance

displacement ...

north is 700+400+100 =1200m n

south=1200m

1200-1200=0


east is 300+300=600m

west is 600m

600-600=0

back at dtart. displ zero


distance is 700+ 300m + 400 m + 600m + 1200m + 300m + 100m  = 3600m


3 0
2 years ago
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