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3 years ago

A possum is trying to cross the street. Its velocity v as a function of time t is given in the graph below

Physics

1 answer:

AlladinOne [14]3 years ago

4 0

Answer:

-10

Explanation:

The rectangular area is the base bbb times height hhh.

\begin{aligned}\Delta x&= A \\\\ &=(2\,\text s)\left (-5\,\dfrac{\text m}{\text s}\right ) \\\\ &= -10\,\text m\end{aligned}

Δx

=(2s)(−5

)

=−10m

The possum displaces 10\,\text m10m10, start text, m, end text to the left.

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A hockey puck on a frozen pond is given an initial speed of 20.0 m/s. If the puck always remains on the ice and slides 115 m bef

bekas [8.4K]

Answer:

μ_k = 0.1773

Explanation:

We are given;

Initial velocity;u = 20 m/s

Final velocity;v = 0 m/s (since it comes to rest)

Distance before coming to rest;s = 115 m

Let's find the acceleration using Newton's second law of motion;

v² = u² + 2as

Making a the subject, we have;

a = (v² - u²)/2s

Plugging relevant values;

a = (0² - 20²)/(2 × 115)

a = -400/230

a = -1.739 m/s²

From the question, the only force acting on the puck in the x direction is the force of friction. Since friction always opposes motion, we see that:

F_k = −ma - - - (1)

We also know that F_k is defined by;

F_k = μ_k•N

Where;

μ_k is coefficient of kinetic friction

N is normal force which is (mg)

Since gravity acts in the negative direction, the normal force will be positive.

Thus;

F_k = μ_k•mg - - - (2)

where g is acceleration due to gravity.

Thus,equating equation 1 and 2,we have;

−ma = μ_k•mg

m will cancel out to give;

-a = μ_k•g

μ_k = -a/g

g has a constant value of 9.81 m/s², so;

μ_k = - (-1.739/9.81)

μ_k = 0.1773

3 0

3 years ago

A 6 N and a 10 N force act on an object. The moment arm of the 6 N force is 0.2 m. If the 10 N force produces five times the tor

Levart [38]

Answer:

The moment arm is 0.6 m

Explanation:

Given that,

First force $F_{1}=6\ N$

Second force $F_{2}=10\ N$

Distance r = 0.2 m

We need to calculate the moment arm

Using formula of torque

$\tau=Force\times lever\ arm$

So, Here,

$\tau_{2}=5 \tau_{1}$

We know that,

The torque is the product of the force and distance.

Put the value of torque in the equation

$F_{2}\times d_{2}=5\times F_{1}\times r_{1}$

$r_{2}=\dfrac{5\times F_{1}\times r_{1}}{F_{2}}$

Where, $F_{1}$ =First force

$F_{1}$ =First force

$F_{2}$ =Second force

$r_{1}$ = distance

Put the value into the formula

$r_{2}=\dfrac{5\times6\times0.2}{10}$

$r_{2}=0.6\ m$

Hence, The moment arm is 0.6 m

6 0

3 years ago

An old millstone, used for grinding grain in a gristmill, is a solid cylindrical wheel that can rotate about its central axle wi

MAXImum [283]

Answer:

The answer is I=70,513kgm^2

Explanation:

Here we will use the rotational mechanics equation T=Ia, where T is the Torque, I is the Moment of Inertia and a is the angular acceleration.

When we speak about Torque it´s basically a Tangencial Force applied over a cylindrical or circular edge. It causes a rotation. In this case, we will have that T=Ft*r, where Ft is the Tangencial Forge and r is the radius

Now we will find the Moment of Inertia this way:

$Ft*r=I*a$ -> $(Ft*r)/(a) = I$

Replacing we get that I is:

$I=(200N*0,33m)/(0,936rad/s^2)$

Then $I=70,513kgm^2$

In case you need to find extra information, keep in mind the Moment of Inertia for a solid cylindrical wheel is:

$I=(1/2)*(m*r^2)$

4 0

3 years ago

A 240 g toy car is placed on a narrow 60-cm-diameter track with wheel grooves that keep the car going in a circle. The 1.0 kg tr

lesya [120]

Answer:

The track's angular velocity is W2 = 4.15 in rpm

Explanation:

Momentum angular can be find

I = m*r^2

P = I*W

So to use the conservation

P1 + P2 = 0

I1*W1 + I2*W2 = 0

Solve to w2 to find the angular velocity

0.240kg*0.30m^2*0.79m/s=-1kg*0.30m^2*W2

W2 = 0.435 rad/s

W2 = 4.15 rpm

8 0

3 years ago

PLEASE HELP THIS IS DUE LIKE HELP ME PLEASE

marysya [2.9K]

Answer:

At the end points of motion (either side) the velocity must be zero because the velocity is changing from - to + (it can't turn around around without passing thru zero,

The velocity will then increase to the midpoint of the motion.

m g h = 1/2 m v^2 where h is the vertical distance thru which the pendulum travels

5 0

2 years ago