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snow_tiger [21]
3 years ago
14

A(n) 1.3 kg mass sliding on a frictionless surface has a velocity of 7.1 m/s east when it undergoes a one-dimensional elastic co

llision with an unknown mass that is initially at rest. After the collision the velocity of the 1.3 kg is 1.7 m/s west. What is the unknown mass? Answer in units of kg
Physics
1 answer:
Oxana [17]3 years ago
3 0

Answer: 2.12 kg

Explanation:

Since the 1.3 kg object moves to the west after the collision, the other object will move to the east after the collision.

In an elastic collision, the relative velocity after the collision is the opposite of the relative velocity before the collision. Since the 1.3 kg object’s velocity before the collision is 6.7 m/s greater than the other object, after the collision, its velocity will be 6.7 m/s less than the other object. To determine the other object’s velocity, use the following equation.

v = 1.7 – 7.1 = -5.4 m/s

The negative sign means it is moving eastward. Let’s use this number is a momentum equation to determine its mass.

Initial momentum = 1.3 * 7.1 = 9.23 east

For the 1.3 object, final momentum = 1.3 * 1.7 = 2.21 west

To determine the final momentum of the other object, add these two numbers.

Final momentum = 11.44 east

To determine its mass, use the following equation.

m * 5.4 = 11.44

m = 11.44 ÷ 5.4 = 2.12 kg

To make sure that kinetic energy is conserved, let’s round this number to 2 kg and determine the final kinetic energies.

For the 1.3 object, KE = 1/2 * 1/3* 1.7^2 = 0.48

For the 2 kg object, KE = 1/2* 2 * 5.4^2 = 29.64

Total final KE = 29.64

Initial KE = 0.5* 1.3 * 7.1^2 = 32.77

Since I rounded the mass up to 2kg, this proves that kinetic energy is conserved and the mass is correct!

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The body is subjected to a force of 0,4 N m with a shoulder of 5 cm. What is the magnitude of this force?​
igor_vitrenko [27]

Answer:

8N

Explanation:

the body is subjected to pressure equals to force force inverse to the area of that body given that it is in motion

5 0
2 years ago
For a short time the missile moves along the parabolic path y=(18−2x2) km. If motion along the ground is measured as x=(4t−3) km
muminat

Answer with Explanation:

We are given that

y=(18-2x^2) km

x=(4t-3)km

Differentiate x and y w.r.t t

\frac{dx}{dt}=4

\frac{dy}{dt}=-4x\frac{dx}{dt}

\frac{dy}{dt}=-4x\times 4=-16x=-16(4t-3)

v_x=\frac{dx}{dt}=4

v_y=\frac{dy}{dt}=-16(4t-3)

Substitute t=1

v_x=4

v_y=-16(4-3)=-16

Magnitude of velocity=\mid v\mid=\sqrt{v^2_x+v^2_y}

\mid v\mid=\sqrt{4^2+(-16)^2}=16.49 m/s

Hence, the magnitude of the missile's velocity=16.49 m/s

a_x=\frac{d(\frac{dx}{dt})}{dt}=\frac{d(4)}{dt}=0

a_y=\frac{d(\frac{dy}{dt})}{dt}=\frac{d(-16(4t-3))}{dt}=-64

Substitute t=1

a_x=0,a_y=-64

\mid a\mid=\sqrt{a^2_x+a^2_y}

\mid a\mid=\sqrt{0+(-64)^2}=64m/s^2

Hence, the magnitude of acceleration when t=1 s=64m/s^2

3 0
3 years ago
9. A wave on Beaver Dam Lake passes by two docks that are 40.0 m apart.
Kobotan [32]

Answers:

a) 10 m

b) time=1.6 s, frquency=0.625 Hz

c) 6.25 m/s

Explanation:

a) If there is a crest at each dock and another three crests between the two docks, and the wavelength \lambda is the distance between to crests; this means we have 4\lambda in 40 m:

40 m=4\lambda

Clearing \lambda:

\lambda=\frac{40 m}{4}

\lambda=10 m

b) This part can be solved by a Rule of Three:

If 10 waves ---- 16 s

1 wave ----- T

Then:

T=\frac{(1 wave)(16 s)}{10 waves}

T=1.6 s This is the period of the wave

On the other hand, the frequency f of the wave has an inverse relation with its period T:

f=\frac{1}{T}

f=\frac{1}{1.6 s}

f=0.625 Hz This is the frequency of the wave

c) The speed v of a wave is given by the following equation:

v=\frac{\lambda}{T}

v=\frac{10 m}{1.6 s}

Finally:

v=6.25 m/s

4 0
3 years ago
Which example nest represents balanced forces ???
n200080 [17]

Answer:

an elevator stopped on the third floor a basketball shot into a hoop a sled sliding down a snowy hill a tow truck pulling a car out of a ditch

5 0
3 years ago
A sphere has surface area 1.25 m2, emissivity 1.0, and temperature 100.0°C. What is the rate at which it radiates heat into empt
Yuri [45]
The total power emitted by an object via radiation is:
P=A\epsilon \sigma T^4
where:
A is the surface of the object (in our problem, A=1.25 m^2
\epsilon is the emissivity of the object (in our problem, \epsilon=1)
\sigma = 5.67 \cdot 10^{-8} W/(m^2 K^4) is the Stefan-Boltzmann constant
T is the absolute temperature of the object, which in our case is T=100^{\circ} C=373 K

Substituting these values, we find the power emitted by radiation:
P=(1.25 m^2)(1.0)(5.67 \cdot 10^{-8}W/(m^2K^4)})(373 K)^4=1371 W = 1.4 kW
So, the correct answer is D.
6 0
3 years ago
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