All categories

2 years ago

A physical quantity, G, is defined by G = (Original mass x time)/(change in mass), what is the S.I. unit of G ?

1 answer:

4 0

The gravitational constant (G) in its base SI units is

3/2
m
3
k
g
/
s
2

But is often seen written as

⋅
N
⋅
2/2
m
2
/
k
g
2

Where N is the Newton unit. N=kg ⋅
⋅
m/s 2
2

You might be interested in

HELP ASAP!!! PLEASE!!

prohojiy [21]

Answer:

Explanation:

The weaker force is pulling to direction East(450 newton)

the net force is (500-450=50 newton) and will move toward west ( in 500 newton direction)

6 0

3 years ago

a hot piece of copper is placed in an insulated cup. what is the final temperature of the water and copper?

slavikrds [6]

Answer:

Option C. 30°C.

Explanation:

The following data were obtained from the question:

Mass of water (Mw) = 0.5 Kg

Specific heat capacity of water (Cw) = 4.18 KJ/Kg°C

Initial temperature of water (Tw1) =

22°C

Change in temperature (ΔT) = T2 – Tw1 = T2 – 22°C

Mass of copper (Mc) = 0.5 Kg

Specific heat capacity of copper (Cc) = 0.386 KJ/kg°C

Initial temperature of copper (Tc1) = 115°C

Change in temperature (ΔT) = T2 – Tc1 = T2 – 115°C

Final temperature (T2) =..?

Note: Both the water and the piece copper will have the same final temperature and the heat will be zero since the water will cool the piece of copper.

Thus, we can determine the final temperature of the water and copper as follow:

Q = MwCwΔT + McCcΔT

0 = 0.5 x 4.18 x (T2 – 22) + 0.5 x 0.386 x (T2 – 115)

0 = 2.09 (T2 – 22) + 0.193 (T2 – 115)

0 = 2.09T2 – 45.98 + 0.193T2 – 22.195

Collect like terms

2.09T2 + 0.193T2 = 45.98 + 22.195

2.283T2 = 68.175

Divide both side by the coefficient of T2 i.e 2.283

T2 = 68.175/2.283

T2 = 29.8 ≈ 30°C

Therefore, the final temperature of water and copper is 30°C.

8 0

3 years ago

A roller coaster car starts from rest at the top of a hill 15 m high and rolls down to ground level. From there it starts into a

Softa [21]

Answer:

955.5N

Explanation:

The normal force is given by the difference between the centripetal force and gravity at the top of the loop:

$F_N = F_C - F_G = m\frac{v^{2} }{r} - mg$

mass m = 65kg

radius of the loop r = 4m

velocity v = ?

g = 9.8 m/s²

To find the centripetal force, you need to find the velocity of the car at the top of the loop.

Use energy conservation:

$E_{tot}=mgh + \frac{1}{2} mv^{2}$

At the top of the hill:

$E_{tot}= mgh_{hill}$

At the top of the loop:

$E_{tot}=mgh_{loo}_p +\frac{1}{2} m v^{2}$

Setting both energies equal and canceling the mass m gives:

$gh_{hill} = gh_{loo}_p + \frac{1}{2} v^{2}$

Solving for v:

$v^{2} = 2g(h_{hill}-h_{loo}_p)$

Using v in the first equation:

$F_N = \frac{2mg(h_{hill}-h_{loo}_p)}{r} - mg$

$F_N = 955.5N$

7 0

3 years ago

What landform is created when magma squeezes between rock layers to form a hardened slab of rock?

Zigmanuir [339]

<span>D. sill

A sill forms when hot magma squeezes between layers of rock to form a hardened sheet of rock. A sill is basically a tabular sheet intrusion between old layers of sedimentary rock, volcanic beds, or in the direction of foliation inherent in metamorphic rock. The word sill means a concordant intrusive sheet.</span>

5 0

3 years ago

Pics pls you know what I mean

V125BC [204]

boiiii I wont show you nun

4 0

3 years ago