Answer:
(a) Amplitude=0.0760 m
(b) Speed=0.337 m/s
Explanation:
(a) For amplitude
We can use the mentioned description of the motion and the energy conservation principle to find amplitude of oscillatory motion
![k_{i}+U_{i}=K_{f}+U_{f}\\ (1/2)mv^{2}+0=0+(1/2)kA^{2}\\ A^{2}=\frac{mv^{2}}{k} \\A=\sqrt{\frac{mv^{2}}{k}}\\ A=\sqrt{\frac{m}{k} }v\\ A=\sqrt{\frac{(0.800kg)}{16N/m} }(0.34m/s)\\A=0.0760m](https://tex.z-dn.net/?f=k_%7Bi%7D%2BU_%7Bi%7D%3DK_%7Bf%7D%2BU_%7Bf%7D%5C%5C%20%281%2F2%29mv%5E%7B2%7D%2B0%3D0%2B%281%2F2%29kA%5E%7B2%7D%5C%5C%20%20A%5E%7B2%7D%3D%5Cfrac%7Bmv%5E%7B2%7D%7D%7Bk%7D%20%5C%5CA%3D%5Csqrt%7B%5Cfrac%7Bmv%5E%7B2%7D%7D%7Bk%7D%7D%5C%5C%20A%3D%5Csqrt%7B%5Cfrac%7Bm%7D%7Bk%7D%20%7Dv%5C%5C%20A%3D%5Csqrt%7B%5Cfrac%7B%280.800kg%29%7D%7B16N%2Fm%7D%20%7D%280.34m%2Fs%29%5C%5CA%3D0.0760m)
(b) For Speed
Again we can use the mentioned description of the motion and the energy conservation principle to find amplitude of oscillatory motion
![k_{i}+U_{i}=K_{f}+U_{f}\\ (1/2)m(v_{i})^{2}+0=(1/2)m(v_{f} )^{2}+(1/2)k(A/2)^{2}\\ (1/2)m(v_{i})^{2}=(1/2)m(v_{f} )^{2}+(1/2)k(A/2)^{2}\\(1/2)m(v_{i})^{2}-(1/2)k(A/2)^{2}=(1/2)m(v_{f} )^{2}\\(1/2)[m(v_{i})^{2}-k(A/2)^{2}]=(1/2)m(v_{f} )^{2}\\(v_{f} )^{2}=1/m[m(v_{i})^{2}-k(A/2)^{2}]\\As\\x=0.250A\\(v_{f} )^{2}=(1/0.800kg)[0.800kg(0.34m/s)^{2}-(16N/m)(0.250(0.07602m)/2)^{2}\\(v_{f} )^{2}=0.1138\\ v_{f}=\sqrt{0.1138}\\ v_{f}=0.337m/s](https://tex.z-dn.net/?f=k_%7Bi%7D%2BU_%7Bi%7D%3DK_%7Bf%7D%2BU_%7Bf%7D%5C%5C%20%281%2F2%29m%28v_%7Bi%7D%29%5E%7B2%7D%2B0%3D%281%2F2%29m%28v_%7Bf%7D%20%29%5E%7B2%7D%2B%281%2F2%29k%28A%2F2%29%5E%7B2%7D%5C%5C%20%281%2F2%29m%28v_%7Bi%7D%29%5E%7B2%7D%3D%281%2F2%29m%28v_%7Bf%7D%20%29%5E%7B2%7D%2B%281%2F2%29k%28A%2F2%29%5E%7B2%7D%5C%5C%281%2F2%29m%28v_%7Bi%7D%29%5E%7B2%7D-%281%2F2%29k%28A%2F2%29%5E%7B2%7D%3D%281%2F2%29m%28v_%7Bf%7D%20%29%5E%7B2%7D%5C%5C%281%2F2%29%5Bm%28v_%7Bi%7D%29%5E%7B2%7D-k%28A%2F2%29%5E%7B2%7D%5D%3D%281%2F2%29m%28v_%7Bf%7D%20%29%5E%7B2%7D%5C%5C%28v_%7Bf%7D%20%29%5E%7B2%7D%3D1%2Fm%5Bm%28v_%7Bi%7D%29%5E%7B2%7D-k%28A%2F2%29%5E%7B2%7D%5D%5C%5CAs%5C%5Cx%3D0.250A%5C%5C%28v_%7Bf%7D%20%29%5E%7B2%7D%3D%281%2F0.800kg%29%5B0.800kg%280.34m%2Fs%29%5E%7B2%7D-%2816N%2Fm%29%280.250%280.07602m%29%2F2%29%5E%7B2%7D%5C%5C%28v_%7Bf%7D%20%29%5E%7B2%7D%3D0.1138%5C%5C%20v_%7Bf%7D%3D%5Csqrt%7B0.1138%7D%5C%5C%20v_%7Bf%7D%3D0.337m%2Fs)
Answer:
1.20372
Explanation: start with 39 times 2 for how much grams each day and then multiply that by 7 then the convert grams into pounds
The answer is the the first one
Answer:
Approximately
if that athlete jumped up at
. (Assuming that
.)
Explanation:
The momentum
of an object is the product of its mass
and its velocity
. That is:
.
Before the jump, the speed of the athlete and the earth would be zero (relative to each other.) That is:
and
. Therefore:
and
.
Assume that there is no force from outside of the earth (and the athlete) acting on the two. Momentum should be conserved at the instant that the athlete jumped up from the earth.
Before the jump, the sum of the momentum of the athlete and the earth was zero. Because momentum is conserved, the sum of the momentum of the two objects after the jump should also be zero. That is:
.
Therefore:
.
.
Rewrite this equation to find an expression for
, the speed of the earth after the jump:
.
The mass of the athlete needs to be calculated from the weight of this athlete. Assume that the gravitational field strength is
.
.
Calculate
using
and
values from the question:
.
The negative sign suggests that the earth would move downwards after the jump. The speed of the motion would be approximately
.
Answer:
<em>The power required to stop the car = 8 × 10⁵ watt.</em>
Explanation:
Kinetic Energy: This is the energy of a body in motion. The S.I unit of energy is Joules (J).
Power: These is defined as time rate of doing work, or it can be defined as the time rate of transfer of energy. The S.I unit of power is Watt (W).
Mathematically, power can be expressed as
Power = Energy/time
P = E/t................. Equation 1
<em>Given: E = 8×10⁶ J, t = 10 s</em>
<em>Substituting these values into equation 1,</em>
<em>p = 8×10⁶/10</em>
<em>p = 8 × 10⁵ watt.</em>
<em>Therefore the power required to stop the car = 8 × 10⁵ watt.</em>