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3 years ago

5.8 meters/second + 1.2 meters/second

Physics

2 answers:

Alex Ar [27]3 years ago

8 0

Answer:7m/s

Explanation:

5.8+1.2

=7.

Natali5045456 [20]3 years ago

4 0

7 i have done this before

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The car travels 25 miles in the first 0.5 hours

Westkost [7]

Answer:

$\huge\boxed{\sf v = 50\ miles / hr}$

Explanation:

<u>Given Data:</u>

Distance = S = 25 miles

Time = t = 0.5 hours

<u>Required:</u>

Speed = v = ?

<u>Formula:</u>

v = S/t

<u>Solution:</u>

v = 25 / 0.5

v = 50 miles / hr

$\rule[225]{225}{2}$

Hope this helped!

<h3>~AnonymousHelper1807</h3>

4 0

3 years ago

An electric current flows through a parallel circuit. Which units tell you about the amount of electrons traveling in the circui

kompoz [17]

'Ampere' is the unit of current. That's the rate at which
electrons travel in the circuit ... the number of electrons
every second. If you wanted the actual amount or number
of electrons, you'd need to know the length of time too.

It doesn't matter whether we're talking about a parallel or
series circuit.

8 0

3 years ago

In which of the following situation is light most likely to be a reflection

maxonik [38]

Answer:

when the reflecting surface is plain and without even small hurdles that are not the visible by our naked eyes. Eg : plain mirror

Explanation:

7 0

3 years ago

A force is applied to a block sliding along a surface (Figure 2). The magnitude of the force is 15 N, and the horizontal compone

Lemur [1.5K]

Answer:

Fy = 14.3 [N]

Explanation:

To be able to solve this problem we must know that the force is a vector and has magnitude and direction, therefore it can be decomposed into the force in the X & y components:

When we have the components on the horizontal and vertical axes we must use the Pythagorean theorem.

$F = \sqrt{F_{x}^{2} +F_{y}^{2} }$

where:

F = 15 [N]

Fx = horizontal component = 4.5 [N]

Fy = vertical component [N]

$15=\sqrt{4.5^{2}+F_{y}^{2}}\\ 15^{2}= (\sqrt{4.5^{2}+F_{y}^{2}})^{2} \\225 = 4.5^{2}+F_{y} ^{2}\\ F_{y}^{2} =225 -4.5^{2}\\ F_{y}^{2}=204.75\\F_{y}=\sqrt{204.75}\\ F_{y}=14.3 [N]$

7 0

3 years ago

Determine the slope of end a of the cantilevered beam. E = 200 gpa and i = 65. 0(106) mm4

DENIUS [597]

For E = 200 gpa and i = 65. 0(106) mm4, the slope of end a of the cantilevered beam is mathematically given as

A=0.0048rads

<h3>What is the slope of end a of the cantilevered beam?</h3>

Generally, the equation for the is mathematically given as

$A=\frac{PL^2}{2EI}+\frac{ML}{EI}$

Therefore

A=\frac{10+10^2+3^2}{2*240*10^9*65*10^6}+\frac{10+10^3*3}{240*10^9*65*10^{-6}}

A=0.00288+0.00192=0.0048rads

A=0.0048rads

In conclusion, the slope is

A=0.0048rads