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andrew11 [14]
3 years ago
11

What will happen if a car experiences a 300 N force to the right from the engine and a separate 150 N force due to friction and

air resistance to the left? Show your work to support your position.
Physics
1 answer:
sveta [45]3 years ago
7 0
If two forces act on an object in different direction with different the resultant force is the difference between them.the object acts in the direction of greater force.
so, 300-150=150N
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What is the value of work done on an object when a
tino4ka555 [31]
W = F * s
Here, F = 50 N
s = 15 m

Substitute their values, 
W = 50 * 15
W = 750 J

In short, Your Answer would be 750 Joules

Hope this helps!
7 0
3 years ago
. At what velocity will the box in Problem 6 be traveling when it hits the ground? Use formula: Va = Vf - Vi / 2
VladimirAG [237]

Answer:

How much time does his victim on the ground below have to move out of harm's way? At what velocity will the safe hit the ground? sownt d-200m. n a = 100/2. Vi-o.

Explanation:

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3 years ago
To calculate acceleration you must know both the objects velocity and_____
amm1812

You need to know the time as well.

8 0
3 years ago
During World War I, the Germans had a gun called Big Bertha that was used to shell Paris. The shell had an initial speed of 2.61
bonufazy [111]

Answer:

The shell hit at a distance of 1.9 x 10² km

The time of flight of the shell was 5.3 x 10² s

Explanation:

The position of the shell is given by the vector "r":

r  = (x0 + v0 * t * cos α ; y0 + v0 * t * sin α + 1/2 g t²)

where:

x0 = initial horizontal position

v0 = magnitude of the initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration of gravity

When the shell hit, the vertical component (ry) of the vector position r is 0. See figure.

Then:

ry = 0 =  y0 + v0 * t * sin α + 1/2 g t²

Since the gun is at the center of our system of reference, y0 and x0 = 0

0 = t (v0 sin α + 1/2 g t)

t= 0 is discarded as solution

v0 sin α + 1/2 g t = 0

t = -2v0 sin α / g

t = (-2 * 2610 m/s * sin 81.9°)/ (-9.8 m/s²) = 5.3 x 10² s. This is the time of flight of the shell until it hit.

Then, the distance at which the shell hit is:

Distance = Module of r = ( x0 + v0 * t * cos α; 0) = x0 + v0 * t * cos α  

Distance = 2.61 km/s * 5.3 x 10² s * cos 81.9 = 1.9 x 10² km

7 0
3 years ago
If m represent mass in kg, v represents speed in m/s and r represents radius in m show F in the formula F= (mv^2)/r can be expre
Dmitrij [34]
M <span>represent mass in kg
</span><span>v represents speed in m/s
</span><span>r represents radius in m

Now, just substitute these into the formula:
</span>F =  \frac{m* v^{2} }{r} =\frac{kg* ( \frac{m}{s} )^{2} }{m} =\frac{kg* \frac{m^{2}}{s^{2}} }{m} = \frac{kg*m^{2}}{s^{2}*m } =\frac{kg*m}{s^{2} }<span>

</span>
3 0
3 years ago
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