Ask question

Ask question

All categories

Arte-miy333 [17]

3 years ago

9

A body is moving vertically upwards. It’s velocity changes at a constant rate from 50m/s to 20m/s in 3 sec. What is it’s acceler

ation ?

2 answers:

sertanlavr [38]3 years ago

3 0

Answer:

-10 m/s²

Explanation:

a = Δv / Δt

a = (20 m/s − 50 m/s) / 3 s

a = -10 m/s²

svp [43]3 years ago

3 0

Answer 10 m/s²

Explanation:

a = Δv / Δt

a = (20 m/s − 50 m/s) / 3 s

a = -10 m/s²

Explanation:

You might be interested in

A vertical steel beam in a building supports a load of 6.0×10⁴. If the length of the beam is 4.0m and it's cross-sectional area

Furkat [3]

Answer:

DL = 1.5*10^-4[m]

Explanation:

First we will determine the initial values of the problem, in this way we have:

F = 60000[N]

L = 4 [m]

A = 0.008 [m^2]

DL = distance of the beam compressed along its length [m]

With the following equation we can find DL

$\frac{F}{A} = Y*\frac{DL}{L} \\where:\\Y = young's modulus = 2*10^{11} [Pa]\\DL=\frac{F*L}{Y*A} \\DL=\frac{60000*4}{2*10^{11} *0.008} \\DL= 1.5*10^{-4} [m]$

Note: The question should be related with the distance, not with the diameter, since the diameter can be found very easily using the equation for a circular area.

$A=\frac{\pi}{4} *D^{2} \\D = \sqrt{\frac{A*4}{\pi} } \\D = \sqrt{\frac{0.008*4}{\\pi } \\\\D = 0.1[m]$

3 0

3 years ago

Objects falling through the air experience a type of friction called air resistance

polet [3.4K]

True is the answer ndndndjjdjd

8 0

2 years ago

Read 2 more answers

9. All of the following are adaptations of herbivores EXCEPT:

postnew [5]

Answer:

B.KEEN EYESIGHT thats for predators !

Explanation:

Herbivores have multiple stomachs to process tough foods, coats to blend in so they don't get eaten, and tough teeth for chewing on the plants. THEY DO NOT HAVE PREY, THEY ARE THE PREY.

5 0

3 years ago

A force Fof 40000 lbf is applied to rod AC the negative Y-direction. The rod is 1000 inches tall. A Force P of 25 lbf is applied

erastova [34]

Answer:

The answer is "effective stress at point B is 7382 ksi "

Explanation:

Calculating the value of Compressive Axial Stress:

$\to \sigma y =\frac{F}{A} = \frac{4 F}{( p d ^2 )} = \frac{(4 x ( - 40000 \ lbf))}{[ p \times (1 \ in)^2 ]} = - 50.9 \ ksi \\$

Calculating Shear Transverse:

$\to \frac{4v}{ 3 A} = \frac{4 (75 \ lbf + 25 \ lbf)}{ \frac{3 ( lni)^2}{4}}$

$= \frac{4 (100 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\ = \frac{400 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\= 0.17 \ ksi$

$= R \times 200 \ in - P \times 100 \ in = 12500 \ lbf \times\ in$

$\to \sigma' =[ s y^2 +3( t \times y^2 + t yz^2 )] \times \frac{1}{2}\\\\$

$= [ (-50.9)^2 +3((63.7)^2 +(0.17)^2 )] \times \frac{1}{2}\\\\=[2590.81+ 3(4057.69)+0.0289]\times \frac{1}{2}\\\\=[2590.81+12,173.07+0.0289] \times \frac{1}{2}\\\\=14763.9089\times \frac{1}{2}\\\\ = 7381.95445 \ ksi\\\\ = 7382 \ ksi$

8 0

3 years ago

10. A force F S 1 of magnitude 6.00 units acts on an object at the ori- gin in a direction u 5 30.0° above the positive x axis (

MArishka [77]

Answer:

The x component of the x axis is: $F_x = F_1\cos(30)=5\cos(30)= 4.33 u$

The y component of the y axis is: $F_y =F_2+ F_1\sin(30)=6.00+5\sin(30)= 8.50 u$

The magnitude is given by: $F = \sqrt{F_x+F_y}= 9.54 u$

The direction is given by: $\theta = \tan^{-1}\frac{F_y}{F_x}= 63.0 ^\circ$

See the graphical magnitude and direction below.

Explanation:

The x component of the x axis is: $F_x = F_1\cos(30)=5\cos(30)= 4.33 u$

The y component of the y axis is: $F_y =F_2+ F_1\sin(30)=6.00+5\sin(30)= 8.50 u$

The magnitude is given by: $F = \sqrt{F_x+F_y}= 9.54 u$

The direction is given by: $\theta = \tan^{-1}\frac{F_y}{F_x}= 63.0 ^\circ$

8 0

3 years ago