All categories

3 years ago

What is the mass of an object if a net force of 8.0 N causes it to accelerate at 1.1 m/s2?

Physics

1 answer:

bija089 [108]3 years ago

6 0

Your answer would be

B. 7.3 KG

You might be interested in

How does an inclined plane change distance and how does it change direction

ElenaW [278]

Objects are known to go down because of a unbalanced force

7 0

3 years ago

PLEASE HELP!! WORTH A TOTAL OF 100 POINTS!!

Serjik [45]

The last equation gives you the tension in the string on the right:

$T_2 = 2Mg - 2Ma = 2M (g-a) = \boxed{33.2 \, \mathrm N}$

4 0

2 years ago

How long would it take an object to stop if it is moving at 15m/s and has an acceleration -0.5m/s^2? Please show your work.

elena-14-01-66 [18.8K]

Time = 15/-0.5 = 14.5/-0.5 = 14.5s

The answer is 14.5s

Sorry no one would answer it sooner.

3 0

3 years ago

A venturi is constructed of a 10.0 cm pipe with a 2.0 cm diameter throat. Water pressure in the pipe is twice atmospheric pressu

postnew [5]

Answer:

$P_t=5066250.0696\ Pa=50\ atm$

Explanation:

Given:

diameter of pipe, $d=0.1\ m$

diameter of throat, $d_t=0.02\ m$

velocity of flow, $v=0.4\ m.s^{-1}$

Pressure in the pipe is twice the atmospheric pressure:

$P=2\times 101325=202650\ Pa$

Now the force of flow of water:

$F=P\times A$

now we find cross sectional area of the pipe:

$A=\frac{\pi.d^2}{4}$

$A=\pi \times\frac{0.1}{4}$

$A=0.007854\ m^2$

Therefore,

$F=202650\times 0.007854$

$F=1591.6094\ N$

Now the area at throat:

$A_t=\frac{\pi.d_t^2}{4}$

$A_t=\frac{\pi\times 0.02^2}{4}$

$A_t=0.000314\ m^2$

Therefore pressure at throat:

$P_t=\frac{F}{A_t}$

$P_t=\frac{1591.6094}{0.000314}$

$P_t=5066250.0696\ Pa=50\ atm$

6 0

3 years ago

Which statements accurately describe the shape of Earth’s orbit around the Sun? Check all that apply.

damaskus [11]

The Earth’s orbit is a nearly circular ellipse.
The Sun is located at one of the two focal points.

The Earth moves around the Sun in an orbit that is almost but not quite circular. As Kepler proved in the seventeenth century, the orbit is actually an ellipse. A parameter called the eccentricity (e) defines the degree of departure from a circle. A value of e=0 would indicate a circle whereas a value of e=0.9 would indicate a very elongated ellipse. The eccentricity of the Earth's orbit is currently e=0.0167.

4 0

3 years ago

Read 2 more answers