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Svetllana [295]
3 years ago
12

two objects of equal mass are a fixed distance apart. if you halve the mass of each object which of the following is most likely

the new gravitational force between them? A half the original force B one fourth of the force C one fourth more than the original force D double the original force
Physics
1 answer:
strojnjashka [21]3 years ago
5 0

Answer:

A) half the original force

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A 111 kg 111 kg horizontal platform is a uniform disk of radius 1.67 m 1.67 m and can rotate about the vertical axis through its
Mashcka [7]

Answer:

287.19 kg.m²

Explanation:

Given that :

The mass M of the horizontal platform  = 111 kg

The radius R of the uniform disk = 1.67 m

mass  of the person standing m_p = 64.7 kg

distance of the person standing d_p = 1.15 m

mass of the dog m_d = 25.7 kg

distance of the dog d_d = 1.35 m

Considering the moment of inertia of the object in the system; the net moment of the inertia can be expressed as:

=  \frac{MR^2}{2}+m_pd_p^2+m_dd_d^2

= (\frac{111*1.67^2}{2})+ (64.7 *1.15^2)+ (25.7*1.35^2)

= 287.19 kg.m²

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4 years ago
When a car traveling at 30m/s hits the gas pedal up to 65m/s in 3.5 sec what is the cars acceleration during that time
Blababa [14]

Answer:

d

Explanation:

4 0
3 years ago
For a birthday gift, you and some friends take a hot-air balloon ride. One friend is late, so the balloon floats a couple of fee
LuckyWell [14K]

Answer:60 kg

Explanation:

Given

mass of people =1140 kg

and balloon is neutral when 1140 kg is loaded in the balloon .

when another person come balloon started descending at the rate of 0.49 m/s^2

For initial condition

F_b=mg

F_b=1140\times 9.8=11.172 KN

For second case

(1140+m)\times 9.8-F_b=(1140+m)\times 0.49

(1140+m)\times (9.8-0.49)=11.172 \times 10^3

1140+m=1200

m=60 kg

5 0
3 years ago
What is the acceleration of a car that slows down from 25.0 m/s to 10.0 m/s in 30 seconds? The car is traveling south.
lorasvet [3.4K]
100 miles per hour ...........
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3 years ago
3. When a magnetic sector instrument was operated with an accelerating voltage of 4.50*103 V, a field of 0.251 T was required to
Damm [24]

Answer:

The answer is "4,500 - 225 \ V".

Explanation:

Using formula for calculating the Voltage:

M_1=12.5\\\\M_2=250\\\\V_1=4,500 \\\\\bold{\text{Formula: }}\\\\\to \bold{\frac{m_1}{m_2}=\frac{V_2}{V_1}}\\\\\to \frac{12.5}{250}=\frac{V_2}{4,500}\\\\\to 0.05=\frac{v_2}{4,500}\\\\\to 0.05\times 4,500= V_2\\\\\to V_2=225\\\\

Hence the range of accelerating in voltage is 4,500 - 225 \ V

8 0
3 years ago
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