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Tamiku [17]
2 years ago
13

Two frictionless lab carts start from rest and are pushed along a level surface by a constant force. Students measure the magnit

ude and duration of the force on each cart, as shown in the partially completed data tableabove, and calculate final kinetic energy and momentum. Which cart has a greater kinetic energy at the end of the push?

Physics
1 answer:
Maru [420]2 years ago
3 0

Answer:

The first cart has greater kinetic energy.

Explanation:

The mass of the second cart is 2kg, and its momentum is 10kg m/s ;therefore, it must be that

(2kg)v = 10kg\: m/s

which gives

v = 5m/s.

Then the kinetic energy of 3kg cart is

K.E.  = \dfrac{1}{2}mv^2

= \dfrac{1}{2}(2kg)(5m/s)^2 =  25J

\boxed{K.E = 25J}

which is less than that of the 1kg cart ; therefore, the 1st cart has greater kinetic energy.

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The temperature and pressure at the surface of Mars during a Martian spring day were determined to be -50 °C and 900 Pa, respect
Sidana [21]

Answer:

T = 273 + (-50) = 273 – 50 = 223 K

R = 188.82 J / kg K for CO2

Density (Martian Atmosphere) = P / RT = 900 / 188.92 x 223 = 900 / 42129.16 = 0.0213 kg / m^{3}

T = 273 +18 = 291 K, R = 287 J / kg k (for air) P = 101.6 k Pa = 101600 Pa

Density (Earth Atmosphere) = P / RT = 101600 / 287 x 291 = 1.216 kg / m^{3}

4 0
2 years ago
Read 2 more answers
A yo-yo can be thought of as a solid cylinder of mass m and radius r that has a light string wrapped around its circumference (s
lbvjy [14]

Answer:

6.5 m/s^2

Explanation:

The net force acting on the yo-yo is

F_net = mg-T

ma=mg-T

now T= mg-ma

net torque acting on the yo-yo is

τ_net = Iα

I= moment of inertia (= 0.5 mr^2 )

α = angular acceleration

τ_net = 0.5mr^2(a/r)

Tr= 0.5mr^2(a/r)

(mg-ma)r=0.5mr^2(a/r)

a(1/2+1)=g

a= 2g/3

a= 2×9.8/3 = 6.5 m/s^2

4 0
2 years ago
The Bernoulli equation is valid for steady, inviscid, incompressible flows with a constant acceleration of gravity. Consider flo
irina1246 [14]

Answer:

p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant

Explanation:

first write the newtons second law:

F_{s}=δma_{s}

Applying bernoulli,s equation as follows:

∑δp+\frac{1}{2} ρδV^{2} +δγz=0\\

Where, δp is the pressure change across the streamline and V is the fluid particle velocity

substitute ρg for {tex]γ[/tex] and g_{0}-cz for g

dp+d(\frac{1}{2}V^{2}+ρ(g_{0}-cz)dz=0

integrating the above equation using limits 1 and 2.

\int\limits^2_1  \, dp +\int\limits^2_1 {(\frac{1}{2}ρV^{2} )} \, +ρ \int\limits^2_1 {(g_{0}-cz )} \,dz=0\\p_{1}^{2}+\frac{1}{2}ρ(V^{2})_{1}^{2}+ρg_{0}z_{1}^{2}-ρc(\frac{z^{2}}{2})_{1}^{2}=0\\p_{2}-p_{1}+\frac{1}{2}ρ(V^{2}_{2}-V^{2}_{1})+ρg_{0}(z_{2}-z_{1})-\frac{1}{2}ρc(z^{2}_{2}-z^{2}_{1})=0\\p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant

there the bernoulli equation for this flow is p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant

note: ρ=density(ρ) in some parts and change(δ) in other parts of this equation. it just doesn't show up as that in formular

4 0
2 years ago
A ball of mass m is found to have a weight Wx on Planet X. Which of the following is a correct expression for the gravitational
Naya [18.7K]

Answer: B. The gravitational field strength of Planet X is Wx/m.

Explanation:

Weight is a force, and as we know by the second Newton's law:

F = m*a

Force equals mass times acceleration.

Then if the weight is:

Wx, and the mass is m, we have the equation:

Wx = m*a

Where in this case, a is the gravitational field strength.

Then, isolating a in that equation we get:

Wx/m = a

Then the correct option is:

B. The gravitational field strength of Planet X is Wx/m.

4 0
2 years ago
A cylindrical specimen of brass that has a diameter listed above, a tensile modulus of 110 GPa, and a Poisson's ratio of 0.35 is
Irina-Kira [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The strain experienced by the specimen is 0.00116 which is option A

Explanation:

The explanation is shown on the second uploaded image

8 0
2 years ago
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