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3 years ago

Zorn and Porsha are ice skating. Porsha has a mass of 60 kg, and Zorn has a mass of 40 kg. As they face each

1 answer:

algol133 years ago

3 0

Assuming the friction between the skaters and the ice is negligible, the magnitude of Porsha's acceleration is 2.8m/s².

Missing part of the question: determine the magnitude of Porsha's acceleration.

Given the data in the question;

Mass of Porsha; $m_{porsha} = 60kg$

Mass of Zorn; $m_{zorn} = 40kg$

Force of Porsha push; $F_{porsha} = 168N$

Magnitude of Porsha's acceleration; $a = \ ?$

To determine the magnitude of Porsha's acceleration, we use Newton's second laws of motion:

$F = m*a$

Where m is the mass of the object and a is the acceleration.

We substitute the mass of Porsha and the force he used into the equation

$168N = 60kg * a\\\\a = \frac{168kg.m/s^2}{60kg}\\\\a = 2.8m/s^2$

Therefore, assuming the friction between the skaters and the ice is negligible, the magnitude of Porsha's acceleration is 2.8m/s².

Learn more: brainly.com/question/25125444

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A meter stick is suspended vertically at a pivot point 22 cm from the top end. It is rotated on the pivot until it is horizontal

suter [353]

Answer:

5.82812 rad/s

Explanation:

L = Length of meter stick = 1 m = 100 cm

$m_c$ = The center of mass of the stick = $\frac{L}{2}-0.22=0.5-0.22=0.28\ m$

$\omega$ = Angular velocity

Moment of inertia of the system is given by

$I=I_c+mr^2\\\Rightarrow I=\frac{mL^2}{12}+mr^2\\\Rightarrow I=\frac{m1^2}{12}+m0.28^2\\\Rightarrow I=m(\frac{1}{12}+0.0784)$

As the energy in the system is conserved

$mgh=I\frac{\omega^2}{2}\\\Rightarrow mgh=m(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow gh=(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow \omega=\sqrt{\frac{2gh}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=\sqrt{\frac{2\times 9.81\times 0.28}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=5.82812\ rad/s$

The maximum angular velocity is 5.82812 rad/s

4 0

3 years ago

Two ice skaters stand in the middle of an ice rink. Drew has a mass of 75 kg, and Lily has a mass of 55 kg. Drew holds Lily, and

ss7ja [257]

PART a)

Before Drew throw Lily in forwards direction they both stays at rest

So initial speed of both of them is zero

So here we can say that initial momentum of both of them is zero

So total momentum of the system initially = ZERO

PART b)

Since there is no external force on the system of two

so there will be no change in the momentum of this system and it will remain same as initial momentum

So final momentum of both of them will be ZERO

PART c)

As we know that momentum of both will be zero always

so we have

$P_1 + P_2 = 0$

$75(v) + 55(2) = 0$

$v = 1.47 m/s$ in opposite direction

7 0

2 years ago

Read 2 more answers

The influence of blood vessel diameter on peripheral resistance is ________.

NeX [460]

Answer:

The influence of diameter of the blood vessel on peripheral resistance is significant because resistance is inversely proportional to the fourth power of the diameter.

Explanation:

The influence of diameter of the blood vessel on peripheral resistance is significant because the relation between the peripheral resistance and the diameter is given as, resistance is inversely proportional to the fourth power of the diameter. Thus, with small increase or decrease in the value of diameter, the peripheral resistance may vary by a significant amount.

6 0

3 years ago

Please Help!

irga5000 [103]

The maximum force that the tires can exert on the road before slipping is 16200 N.

From the information in the question;

The coefficient of static friction = 0.9

The mass of the car = 1800 kg

Using the formula;

μ = F/R

μ = coefficient of static friction

F = force on the tires

R = the reaction force

But recall that the reaction is equal in magnitude to the weight of the car.

W=R

Hence; R = 1800 kg × 10 ms-2 = 18000 N

Making F the subject of the formula;

F = μR

Substituting values;

F = 18000 N × 0.9

F = 16200 N

Hence, the maximum force that the tires can exert on the road before slipping is 16200 N.

Learn more: brainly.com/question/18754989

6 0

2 years ago

Which one does not affect the gravitational force between the 1 point

Zinaida [17]

Answer:

D. The net electrical charge of the Earth.

Explanation:

All the other three choices affect the gravitational force. I remember from middle school.

6 0

3 years ago