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3 years ago

A 23 kg child is riding a 6.7 kg bike with a velocity of 4.4 m/s to the northwest.

Physics

1 answer:

mash [69]3 years ago

5 0

How am I supposed to know the momentum of anything ? What does it matter

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What is the equation for an inelastic collision

abruzzese [7]

M1 v1 = (m1 + m2)v2.

All of the exponents should be lowered to the bottom right of the letters.

7 0

3 years ago

A bug splats against the windshield of a car traveling at high speeds down a backcountry road. Which statement correctly compare

zvonat [6]

Answer:

C. The bug's change in momentum is equal to the car's change in momentum.

Explanation:

As we know by Newton's 2nd law

$F = \frac{\Delta P}{\Delta t}$

here we have also know that when car hits the bug then force applied by wind shield on the bug is same as the force applied by the bug on the car's wind shield as per Newton's III law

$F_{12} = F_{21}$

so we know that

$\frac{\Delta P_{12}}{\Delta t} = \frac{\Delta P_{21}}{\Delta t}$

so we have

$\Delta P_{12} = \Delta P_{21}$

so correct answer will be

C. The bug's change in momentum is equal to the car's change in momentum.

6 0

4 years ago

A toaster draws 8 A of current with a voltage of 120 V. Which is the power used by the toaster?

dexar [7]

$U=120 \text{ V}\\
I=8\text{ A}\\
P=U\cdot I\\\\
P=120\cdot8=960 \text{ W}
$

7 0

3 years ago

Question 33

pantera1 [17]

Answer:

A. The waves in the water travel faster and at a higher frequency than they travel on land.

Explanation:

The main reason why human ears can hear dolphins' vocalizations while under the water but cannot hear them well on land is because water is denser than air and air particles travel faster in denser particles.

Denser particles also ensures that the frequency of the waves move faster which in turn produces a faster and louder result.

3 0

3 years ago

Determine the velocity that a car should have while traveling around a frictionless curve of radius 100m and that is banked 20 d

alex41 [277]

Answer:

$v=18.89\frac{m}{s}$

Explanation:

From the free-body diagram for the car, we have that the normal force has a vertical component and a horizontal component, and this component act as the centripetal force on the car:

$\sum F_x:Nsin(20^\circ)=ma_c(1)\\\sum F_y:Ncos(20^\circ)=mg(2)$

Solving N from (2) and replacing in (1):

$N=\frac{mg}{cos(20^\circ)}\\(\frac{mg}{cos(20^\circ)})sin(20^\circ)=ma_c\\gtan(20^\circ)=a_c$

The centripetal acceleration is given by:

$a_c=\frac{v^2}{r}$

Replacing and solving for v:

$gtan(20^\circ)=\frac{v^2}{r}\\v=\sqrt{grtan(20^\circ)}\\v=\sqrt{9.8\frac{m}{s^2}(100m)tan(20^\circ)}\\v=18.89\frac{m}{s}$

4 0

3 years ago