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Sergio [31]
3 years ago
12

Can you please help me with #9?

Physics
1 answer:
IRINA_888 [86]3 years ago
3 0
The answer is 1 m/s. :::))))
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Any help please, physics is not my strong suit?
frez [133]

Answer:

2000mg = 2g

5L = 5000mL

16cm = 160mm

5 0
3 years ago
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A motorcyclist drives around a bend with a 20 m radius, with a constant velocity of 3 m/s. The motorcyclist and the motorcycle h
Pavel [41]

Answer:

a=0.45\ m/s^2

Explanation:

Given that,

The radius of a bend, r = 20 m

Velocity of motorcyclist, v = 3 m/s

The combined mass of motorcyclist and the motorcycle is 50 kg

We need to find the motorcyclist’s centripetal acceleration. The formula used to find the centripetal acceleration is given by :

a=\dfrac{v^2}{r}\\\\a=\dfrac{(3)^2}{20}\\\\a=0.45\ m/s^2

So, the acceleration of the motorcyclist is 0.45\ m/s^2.

3 0
3 years ago
When a man travel from hilly to terai what will happen to weight and why?
juin [17]

Answer:

When a man travels from Hilly region to Terai region, his weight gradually increases because the value of g is more at the Terai region than that in hilly region. 3. An object weights 20 N in air and 16 N in liquid, then answer the following questions.

Explanation:

because the value of g is more at the Terai region than that in hilly region. 3. An object weights 20 N in air and 16 N in liquid, then answer the following questions.

6 0
3 years ago
A bicycle wheel with radius 0.3 m rotates from rest to 3 rev/s in 5 s. What is the magnitude and direction of the total accelera
AlekseyPX

Answer:

Explanation:

Given

Radius of bicycle wheel r=0.3\ m

Initial angular velocity \omega _0=0

It rotates 3 revolution in 5 s therefore

\omega =2\pi 3=\6\pi =18.85\ rad/s

using \omega =\omega _0+\alpha t

where \alpha =angular\ acceleration

\omega =Final\ angular\ velocity

t=time

\alpha =\frac{18.85}{5}=3.77 rad/s^2

Total acceleration of any point will be a vector sum of tangential acceleration and centripetal acceleration

\omega at t=1

\omega =0+3.77\times 1=3.77 rad/s

a_c=\omega ^2\cdot r

a_c=(3.77)^2\cdot 0.3=4.26 m/s^2

Tangential acceleration a_t=\alpha \times r

a_t=3.77\times 0.3=1.13 m/s^2

a_{net}=\sqrt{a_t^2+a_c^2}

a_{net}=\sqrt{(1.13)^2+(4.26)^2}

a_{net}=4.41 m/s^2

                       

7 0
3 years ago
What is the unit of G in the F=Gm1m2/r^2​
kobusy [5.1K]


G
has the SI units
m
3
k
g
⋅
s
2
6 0
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