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3 years ago

A 1.2 x 103 kg racecar, with a velocity of 8 m/s, collides with an unsuspecting 80 kg honey badger who is standing

Physics

1 answer:

aalyn [17]3 years ago

4 0

Answer: 90 m/s

Explanation:

Given

mass of racecar $M=1.2\times10^3\ kg$

velocity of racecar $u=8\ m/s$

mass of still honeybadger $m=80\ kg$

after collision race car is traveling at a speed of $v_1=2\ m/s$

conserving linear momentum

$Mu+m\times0=Mv_1+ mv_2\quad[v_2=\text{velocity of honeybadger after colllision}]$

$1.2\times10^3\times8+0=1.2\times10^3\times2+80\times v_2$

$1.2\times10^3(8-2)=80v_2\\v_2=\frac{7.2\times10^3}{80}=90\ m/s$

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A ball was dropped from a height of 10 feet. Each time it hits the ground, it bounces 4/5 of its previous height. Find the total

Shtirlitz [24]

Answer:

d = 90 ft

Explanation:

As we know that after each bounce it reaches to 4/5 times of initial height

so we can say

$h_2 = \frac{4}{5}h$

so the distance covered is given as

$d = h + 2(\frac{4}{5}h) + 2(\frac{4}{5})^2h + 2(\frac{4}{5})^3h........$

here we know that

h = 10 feet

$d = h + 2(\frac{4}{5}h)(1 + \frac{4}{5} + (\frac{4}{5})^2 + ...........)$

$d = 10 + 2(\frac{4}{5}(10))(\frac{1}{1 - \frac{4}{5}})$

$d = 90 ft$

8 0

3 years ago

A heat engine has a maximum possible efficiency of 0.780. If it operates between a deep lake with a constant temperature of-24.8

Volgvan

Answer : The temperature of the hot reservoir (in Kelvins) is 1128.18 K

Explanation :

Efficiency of carnot heat engine : It is the ratio of work done by the system to the system to the amount of heat transferred to the system at the higher temperature.

Formula used for efficiency of the heat engine.

$\eta =1-\frac{T_c}{T_h}$

where,

$\eta$ = efficiency = 0.780

$T_h$ = Temperature of hot reservoir = ?

$T_c$ = Temperature of cold reservoir = $-24.8^oC=273+(-24.8)=248.2K$

Now put all the given values in the above expression, we get:

$\eta =1-\frac{T_c}{T_h}$

$0.780=1-\frac{248.2K}{T_h}$

$T_h=1128.18K$

Therefore, the temperature of the hot reservoir (in Kelvins) is 1128.18 K

8 0

3 years ago

A 392 N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 24

alex41 [277]

Answer:

h=12.41m

Explanation:

N=392

r=0.6m

w=24 rad/s

$I=0.8*m*r^{2}$

So the weight of the wheel is the force N divide on the gravity and also can find momentum of inertia to determine the kinetic energy at motion

$N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}$

$m=40kg$

moment of inertia

$M_{I}=0.8*40.0kg*(0.6m} )^{2}\\M_{I}=11.5 kg*m^{2}$

Kinetic energy of the rotation motion

$K_{r}=\frac{1}{2}*I*W^{2}\\K_{r}=\frac{1}{2}*11.52kg*m^{2}*(24\frac{rad}{s})^{2}\\K_{r}=3317.76J$

Kinetic energy translational

$K_{t}=\frac{1}{2}*m*v^{2}\\v=w*r\\v=24rad/s*0.6m=14.4 \frac{m}{s}\\K_{t}=\frac{1}{2}*40kg*(14.4\frac{m}{s})^{2}\\K_{t}=4147.2J$

Total kinetic energy

$K=3317.79J+4147.2J\\K=7464.99J$

Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height

$K-W=E_{p}\\7464.99-2600J=m*g*h\\4864.99J=m*g*h\\h=\frac{4864.99J}{m*g}\\h=\frac{4864.99J}{392N}\\h=12.41m$

6 0

3 years ago

The roller coaster is designed according to safety regulations that prohibit the speed of the car from exceeding 20 m/s. find th

hram777 [196]

maximum allowed value of the speed in roller coaster is given as

$v = 20 m/s$

now from kinematics we can say

$v^2 - v_i^2 = 2 a s$

here initial speed will be

$v_i = 0$

acceleration is due to gravity

$a = 9.8 m/s^2$

now we can use this to find the height

$20^2 - 0^2 = 2 * 9.8* h$

$400 = 19.6 *h$

$h = 20.4 m$

so maximum allowed height will be 20.4 m

4 0

3 years ago

Can anyone help me with 11 and 12 please

dexar [7]

The 1st one is basically B because it will stay in motion with the same speed and in the same direction unless acted on by an unbalanced force and the 2cd one is A because most of is transformed into thermos energy. hope this helps!

7 0

4 years ago