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Marina86 [1]
3 years ago
14

A person is pushing a box. The net external force on the 60-kg box is stated to be 90 N. If the force of friction opposing the m

otion is 30 N, what is the acceleration of the box?a. 0.66 m/s2b. 1.5 m/s2c. 2.0 m/s2
Physics
1 answer:
Tems11 [23]3 years ago
5 0

Answer:

b.1.5m/s^2

Explanation:

We are given that

Mass of box=60kg

Net external force applied on the box=90 N

Friction force =30N

We have to find the acceleration of the box.

We know that

Net external force=ma

Substitute the values then we get

90=60a

a=\frac{90}{60}=1.5m/s^2

Hence, option b is true.

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Q. A mass of 300g is lifted to a<br> height of 10m<br> 205 by a person. Calculate his work done
sergiy2304 [10]
A=mgh
m=300g=0.3kg
g=9,81 m/s^2
h=10m
A=29.43J
3 0
2 years ago
Two cars traveled equal distances in different amounts of time. Car A traveled the distance in 2 h, and Car B traveled the dista
lions [1.4K]
The answer is 60 mph.

The speed (v) is distance (d) per time (t): v = d/t

Car A:
v1 = ?
t1 = 2 h
d1 = ?
___
v1 = d1/t1
d1 = v1 * t1

Car B:
v2 = ?
t2 = 1.5 h
d2 = ?
___
v2 = d2/t2
d2 = v2 * t2

<span>Two cars traveled equal distances:
d1 = d2
</span>v1 * t1 = v2 * t2

<span>Car B traveled 15 mph faster than Car A:
v2 = v1 + 15


</span>v1 * t1 = v2 * t2
v2 = v1 + 15
________
v1 * 2 = (v1 + 15) * 1.5
2v1 = 1.5v1 + 22.5
2v1 - 1.5v1 = 22.5
0.5v1 = 22.5
v1 = 22.5/0.5
v1 = 45 mph


v2 = v1 + 15
v2 = 45 + 15
v2 = 60 mph
8 0
3 years ago
wo parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the magnitude
melisa1 [442]

Answer:

<em> -18896.49 V/m</em>

<em></em>

Explanation:

Distance between the two plates = 10 cm = 10 x 10^{-2} m = 0.1 m

Also, one of the plates is taken as<em> zero volt.</em>

a. The potential strength between the zero volt plate, and 7.05 cm (0.0705 m) away is 393 V

b. The potential strength between the other plate, and 2.95 cm (0.0295 m) away is 393 V

<em>Potential field strength = -dV/dx</em>

where dV is voltage difference between these points,

dx is the difference in distance between these points

For the first case above,

potential field strength = -393/0.0705 = -5574.46 V/m

For the second case ,

potential field strength = -393/0.0295 = -13322.03 V/m

Magnitude of the field strength across the plates will be

-5574.46 + (-13322.03) = -5574.46 + 13322.03 =<em> -18896.49 V/m</em>

6 0
3 years ago
What is the efficiency (w/qh) of an ideal carnot heat engine operating between a hot region at t= 400 k and a cold one at t= 300
Vinvika [58]
The efficiency of an ideal Carnot heat engine can be written as:
\eta = 1-  \frac{T_{cold}}{T_{hot}}
where
T_{cold} is the temperature of the cold region
T_{hot} is the temperature of the hot region

For the engine in our problem, we have T_{cold}=300 K and T_{hot}=400 K, so the efficiency is
\eta= 1 - \frac{300 K}{400 K}=0.25
4 0
3 years ago
HELP...<br> 3.00 amu = _____ Mev.<br><br> 3.22 x 10-3<br> 2.79 x 103<br> 3.10 x 102
DanielleElmas [232]
The correct answer is:
2.79 \cdot 10^3 MeV
Let's see why.

1 amu corresponds to the mass of the proton, which is:
m_p = 1.66 \cdot 10^{-27} kg
if we convert this into energy, using Einstein equivalence between mass and energy, we find:
E=mc^2 = (1.66 \cdot 10^{-27} kg)(3\cdot 10^8 m/s)^2 = 1.49 \cdot 10^{-10} J
Now we can convert it into electronvolts:
E= \frac{1.49 \cdot 10^{-10}kg}{1.6 \cdot 10^{-19} J/eV} =9.34 \cdot 10^9 eV = 934 MeV

So, 1 amu = 934 MeV. Therefore, 3 amu corresponds to 3 times this value:
3 amu = 3 \cdot 934 MeV  \sim 2790 MeV = 2.79 \cdot 10^3 MeV
5 0
3 years ago
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