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Goshia [24]
3 years ago
12

Which statement about gravitational attraction is correct?

Physics
1 answer:
natita [175]3 years ago
3 0
It’s b all objects with mass experience gravitational attraction I hope this helps
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Question 9
egoroff_w [7]

Answer:

C

Explanation:

F=ma

given solution

v=12m/s a=v/t

s=6 sec =12m/s÷6sec

=2m/s^2 then we get acceleration now we will find the mass. first derive the the formula of mass by crisis cross then you will get this formula which is m=F/a

=36÷2

= 18

6 0
3 years ago
What does the stirrup press against
AveGali [126]
We need a system to use those air vibrations to push against the surface of the inner ear fluid.
7 0
2 years ago
A chemical lab isolates a gaseous element from the atmosphere that does not form chemical compounds. In which group of the perio
Sholpan [36]

Answer:

A- 18

Explanation:

It wont let me explain how

8 0
3 years ago
A heat-conducting rod consists of an aluminum section, 0.30 m long, and a copper section, .70m long. Both sections have a cross-
Igoryamba
Thermal conductions
K= QL/ART
Aluminium T₁ = 10 + 273.15
                    T₂ = 283.15k
205 = 2.0  × 0.30/4× 10⁻⁴ × (T₂ - 283.15)
Copper
385 = Q × 0.70/4×10⁻⁴ ×(433.15 - T₂)
Where T₃ = 160 + 273.15
T₃ = 433.15K
From 2 to 3
205/385 = 0.30/0.70 × 433.15 - T₂/T₂ - 283.15
= 0.53T₂ -150.06 = 181.92 - 0.42 T₂
→ 0.95T₂ = 331.98 ⇒ T₂ = ₂349.45k
T₂ = 76.3°c
=77°c.

6 0
3 years ago
What is the fundamental frequency (in Hz) of a 0.632 m long tube, open at both ends, on a day when the speed of sound is 344 m/s
Greeley [361]

Answer:

f=272.15Hz

Explanation:

Given data

Length of tube L=0.632 m

Speed of sound v=344 m/s

To find

Fundamental frequency f

Solution

The fundamental frequency of the tube can be given as:

f=\frac{v}{2L}\\ f=\frac{344m/s}{2(0.632m)}\\ f=272.15Hz

4 0
3 years ago
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