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12345 [234]
3 years ago
5

Slow-moving vehicles are prohibited from traveling on expressways or on roadways with a minimum posted speed limit greater than

______and should use the ______ lane of travel on divided highways.
Physics
1 answer:
Bumek [7]3 years ago
7 0
<span>Slow-moving vehicles are prohibited from traveling on expressways or on roadways with a minimum posted speed limit greater than 40mph, and should use the right lane of travel on divided highways.
As slow moving vehicles should use the right lane, fast moving vehicles should use the left lane.
Before start driving, we should learn all the guidelines and rules for this, to avoid any type of inconvenience.</span>
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A 0.50 kg object is at rest. A 2.88 N force to the right acts on the object during a time interval of 1.48 s. a) What is the vel
Murljashka [212]

Answer:

8.5m/s

Explanation:

We are given that

Mass of object=m=0.50 kg

Initial velocity, u=0

Force=F=2.88 N

Time=1.48 s

a.We know that

F=ma

Using the formula

2.88=0.50a

a=\frac{2.88}{0.50}=5.76m/s^2

a=\frac{v-u}{t}

Using the formula

5.76=\frac{v-0}{1.48}

v=5.76\times 1.48=8.5m/s

Hence, the velocity of the object at the end of this time interval=8.5m/s

8 0
3 years ago
WILL GIVE BRAINLIEST HELP FASTWhat is the mass of Planet X? Note: The constant of universal gravity (G) equals 6.674 X 10-11 N·m
disa [49]

Answer:

Explanation:

Centripetal acceleration's equation is:

a_c=\frac{v^2}{r} where v is the velocity of the object (moon II) and r is the radius. We have the radius, but we don't have the velocity, and we can't solve for acceleration until we do have it. Assuming moon II is a circle, or close enough to be called a circle, it has a circumference.

C = 2πr. If we can find the circumference of the circle, we can plug in the orbital period for the time, the circumference for the distance, and solve for velocity in d = rt. So let's do that and see what happens.

C = 2(3.14)(9.0 × 10⁷) and

C = d = 5.7 × 10⁸. Plugging in and solving for v:

5.7*10^8=v(3.0*10^5) and

v = 1.9 × 10³. That is the velocity we can use in the centripetal acceleration equation.

a_c=\frac{(1.9*10^3)^2}{9.0*10^7} and

a_c=.040\frac{m}{s^2}

These are fun!

8 0
2 years ago
A 2.50 gram rectangular object has measurements of 22.0 mm, 13.5 mm, and 12.5 mm. what is the object's density in units of g/ml?
Zinaida [17]
G/mL is equivalent to g/cm^3, so we first convert the dimensions into cm:
2.20 cm, 1.35 cm, and 1.25 cm
Then the total volume is: V = lwh = 3.7125 cm^3
To get the density, we divide mass by volume: 2.50 g / 3.7125 cm^3 = 0.6734 g/cm^3 = 0.6734 g/mL
4 0
3 years ago
If the potential across two parallel plates, separated by 4.0 cm, is 15.0 V, what is the electric field strength in volts per me
abruzzese [7]

Field strength = (15 V) / (4 cm)

Field strength = (15 V) / (0.04 meter)

Field strength = (15/0.04) (volts/meter)

<em>Field strength = 375 volts/meter </em>

3 0
3 years ago
Read 2 more answers
Balance the equation-<br> Al+Mn02 ———-&gt; Mn + Al2O3
joja [24]

Answer:

                                                                                                         

Explanation:

3 0
3 years ago
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