100 POINTS!!!

Find acceleration. Will give brainliest!

zalisa [80]

Answer:

16200 km/s

270 km/min

4.5 km/h

Explanation:

Acceleration Formula: Average Acceleration = Δv/Δt (change in velocity over change in time)

Simply plug in our known variables and solve:

a = (45.0 - 0)/10

a = 45.0/10

a = 4.5 km/h

7 0

3 years ago

Read 2 more answers

In the design of wall and column forms, the two most important factors are which of the following? a. rate of placement of the c

Simora [160]

Answer:

c. length of the wall or column and the rate of placement of the concrete

Explanation:

when designing for wall and column form-works, it is of utmost important to know the length of the wall and the type of concrete placement to be used.

Concrete placement has methods and precaution to be taken when doing the form work

if the concrete placement is manually (hand or funnel) the form work height should not be more than 1 m to enable easy compaction and vibration of concrete in the form.

Also, if the form work length is too long and it is not well reinforced, it tends to burg if the force apply during concrete placement or during vibration is much.

5 0

3 years ago

Starting velocity: 50 m/s

Taya2010 [7]

Please find attached photograph for your answer. Please do comment whether it is useful or not

6 0

3 years ago

The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in m

BARSIC [14]

A) The resultant force is 30.4 N at $25.3^{\circ}$

B) The resultant force is 18.7 N at $43.9^{\circ}$

Explanation:

A)

In order to find the resultant of the two forces, we must resolve each force along the x- and y- direction, and then add the components along each direction to find the components of the resultant.

The two forces are:

$F_1 = 20 N$ at $0^{\circ}$ above x-axis

$F_2 = 15 N$ at $60^{\circ}$ above y-axis

Resolving each force:

$F_{1x}=F_1 cos \theta = (20)(cos 0)=20 N\\F_{1y}=F_1 sin \theta =(20)(sin 0)=0 N$

$F_{2x}=F_2 cos \theta = (15)(cos 60)=7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 60)=13.0 N$

So, the components of the resultant are:

$F_x = F_{1x}+F_{2x}=20+7.5 = 27.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude of the resultant is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{27.5^2+13.0^2}=30.4 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{27.5})=25.3^{\circ}$

B)

In this case, the 15 N is applied in the opposite direction to the 20 N force. Therefore we need to re-calculate its components, keeping in mind that the angle of the 15 N force this time is

$\theta=180^{\circ}-60^{\circ}=120^{\circ}$