A 1000-kg sports car accelerates from rest to 95 km/h in 6.4 s . What is the average power delivered by the engine

The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2.00\

saw5 [17]

Answer:

Explanation:

From the given information:

The torque produced due to the force can be expressed as:

$\tau = F \times r$

where;

$\tau$ = torque

F = force exerted

r = lever's arm radius

$\tau$ = $2.00 \times 10^3 \times 0.03 m$

$\tau$ = 60 N.m

However, equating the torque with the moment of inertia & angular acceleration, we use the equation:

$\tau$ = I∝

60 Nm = I × 120 rad/s²

I = 60 Nm/120 rad/s²

I = 0.5 kg.m²

3 0

3 years ago

WITHOUT FRICTION, what happens to the skater if you release him from a height ABOVE the track?

ElenaW [278]

Answer:

B, he will fly off the other end

Explanation:

there is too much potential energy that transforms into kinetic energy for him to stay on the track.

3 0

3 years ago

Bacteria that can grow in your breathing area is ?

andre [41]

<span>The bacteria is called bacteria pericarditis.

</span>

6 0

3 years ago

Read 2 more answers

A passenger on an interplanetary express bus traveling at = 0.95 takes a 9.0-minute catnap, according to her watch.How long does

Nimfa-mama [501]

Answer:

28.82 minutes

Explanation:

This problem is an example of time dilatation by acceleration. The formula is:

$\Delta t=\gamma \Delta t_0=\frac{\Delta t_0}{\sqrt{1-\frac{v^2}{c^2} }}$

$\Delta t$ is the time measured by the observer on a fixed position (in this case in the fixed planet), $\Delta t_0$ is the time measured by the person moving away from the fixed observer (the one on an interplanetary express bus), $c$ is the speed of light and $v$ the velocity of the observer that is moving away (the velocity of the ship).

$\Delta t_0=9 min=540s\\\Delta t=\frac{540}{\sqrt{1-\frac{(0.95c)^2}{c^2}}}=\frac{540}{\sqrt{1-\frac{0.95^2c^2}{c^2}}}=\frac{540}{\sqrt{1-0.95^2}}\\\Delta t=\frac{540}{\sqrt{1-0.9025}}=\frac{540}{\sqrt{0.0975}}=\frac{540}{0.3122}=1729.38s\\\Delta t=28.82minutes$

6 0

4 years ago

The inventor of a new machine claims that its actual mechanical advantage is 4. Literature on the machine reports its input dist

Andreas93 [3]

To solve this we are going to use the formula for ideal mechanical advantage: $IMA= \frac{D_{I} }{D_{O}}$
where
$IMA$ is the machine mechanical advantage
$D_{I}$ is the input distance
$D_{O}$ is the output distance

We know for our problem that $D_{I}=81$ and $D_{O}=27$ . Lets replace those values in our formula to find $IMA$ :
$IMA= \frac{D_{I} }{D_{O}}$
$IMA= \frac{81}{27}$
$IMA=3$

The ideal machine advantage of the machine is 3. The inventor is claiming that the actual mechanical advantage of the machine is 4. Since the actual mechanical advantage takes into account energy losses, it is always less than the ideal mechanical advantage. We can conclude that the developer's claim is false.

8 0

3 years ago