All categories

3 years ago

Is it true to say there are forces acting on an object that is traveling at 20 m/s in a

Physics

2 answers:

notsponge [240]3 years ago

7 0

Answer:

<em>Its true</em> i think

Explanation:

natulia [17]3 years ago

7 0

There don’t have to be forced acting on something that is travelling at a constant velocity (Newton’s 1st Law).

If forced are acting they must be balanced forced with Net force = zero.

You might be interested in

Can some one pls answer them asapppp

Yuliya22 [10]

A) Oil B) Wood C) 0.02J D) 3g E) Gas

4 0

2 years ago

Read 2 more answers

PLEASE HELP!! 13 POINTS! SCIENCE!

pogonyaev

Answer:

Explanation:

3 0

3 years ago

Read 2 more answers

Consider a plywood square mounted on an axis that is perpendicular to the plane of the square and passes through the center of t

belka [17]

Answer:

T= 8.061N*m

Explanation:

The first thing to do is assume that the force is tangential to the square, so the torque is calculated as:

T = Fr

where F is the force, r the radius.

if we need the maximum torque we need the maximum radius, it means tha the radius is going to be the edge of the square.

Then, r is the distance between the edge and the center, so using the pythagorean theorem, r i equal to:

r = $\sqrt{(0.38m)^2+(0.38m)^2}$

r = 0.5374m

Finally, replacing the value of r and F, we get that the maximun torque is:

T = 15N(0.5374m)

T= 8.061N*m

4 0

3 years ago

A war wolf is a device used during the middle ages to assault fortifications with large rocks. A simple trebuchet is constructed

Katarina [22]

Answer:v=41.23 m/s

Explanation:

Given

mass of heavy object $m_1=52 kg$

distance of $m_1$ from the axle $r_1=14 cm$

mass of rock $m_2=123 gm$

Length of rod $=4.1 m$

distance of $m_2$ from axle $r_2=4.1-0.14=3.96 m$

Net torque acting is

$T_{net}=m_1gr_1-m_2gr_2$

$T_{net}=52\times 0.14\times g-0.123\times 3.96\times g$

$T_{net}=6.793\times 9.8$

$T_{net}=66.57 N-m$

Work done by $T_{net}$ is converted to rock kinetic Energy

thus

$T_{net}\times \theta =\frac{mv^2}{2}$

Where $\theta =angle\ turned =\frac{\pi }{2}$

$v= velocity\ at\ launch$

$66.57\times \frac{\pi }{2}=\frac{0.123\times v^2}{2}$

$v^2=66.57\times \pi$

$v=\sqrt{1700.511}$

$v=41.23 m/s$

3 0

3 years ago

Define limitations in the operation conditions of a pn junction<br>

suter [353]

Answer:

Such limitations are given below.

Explanation:

Each pn junction provides limited measurements of maximum forwarding current, highest possible inversion voltage as well as the maximum output level.
If controlled within certain adsorption conditions, the pn junction could very well offer satisfying performance.
In connector operation, the maximum inversion voltage seems to be of significant importance.

6 0

3 years ago