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3 years ago

When the mass of an object decreases, the force of gravity - Remains Unchanged - Decreases - increases - Becomes irregular

Physics

2 answers:

Gwar [14]3 years ago

5 0

The force of gravity decreases

vekshin13 years ago

3 0

Answer:

Decreases

Explanation More mass=More gravitational pull

Less mass=Less gravitational pull

Hope this helps

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A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 19.0 m/s when the

german

Answer: 4 s

Explanation:

Given

The ball leaves the hand of student with a speed of $u=19\ m/s$

When the hand is $h=2.5\ m$ above the ground

Using the equation of motion we can write

$h=ut+\dfrac{1}{2}at^2$

Substitute the values

$\Rightarrow 2.5=-19t+0.5\times 9.8t^2\\\Rightarrow 4.9t^2-19t-2.5=0\\\\\Rightarrow t=\dfrac{19\pm \sqrt{(-19)^2-4\times 4.9\times (-2.5)}}{2\times 19}\\\Rightarrow t=4.0049\quad [\text{Neglecting the negative value of }t]$

Thus, the ball will take 4 s to hit the ground.

5 0

2 years ago

Direct tension indicators are sometimes used instead of torque wrenches to ensure that a bolt has a prescribed tension when used

natali 33 [55]

Complete Question

The complete question is shown on the uploaded image

Answer:

The tension on the shank is $T =8391.6 N$

Explanation:

From the question we are told that

The strain on the strain on the head is $\Delta l = 0.1 mm/mm = \frac{0.1}{1000} = 0.1 *10^{-3} m/m$

The contact area is $A = 2.8 mm^2 = 2.8* (\frac{1}{1000} )^2 = 2.8*10^{-6} m^2$

Looking at the first diagram

At 600 MPa of stress

The strain is $0.3mm/mm$

At 450 MPa of stress

The strain is $0.0015 mm/mm$

To find the stress at $\Delta l$ we use the interpolation method

$\frac{\sigma_{\Delta l} - \sigma_{0.0015} }{ \sigma _ {0.3} - \sigma_{0.0015} } = \frac{e_{\Delta l } - e_{0.0015}}{e_{0.3} - e_{ 0.0015}}$

Substituting values

$\frac{\sigma _{\Delta l} - 450}{600 - 450} = \frac{0.1 -0.0015}{0.3 - 0.0015}$

$\sigma _{\Delta l} -450 = 49.50$

$\sigma _{\Delta l} = 499.50 MPa$

Generally the force on each head is mathematically represented as

$F = \sigma_{\Delta l} * A$

Substituting values

$F = 499.50*10^{6} * 2.8*10^{-6}$

$=1398.6N$

Now the tension on the bolt shank is as a result of the force on the 6 head which is mathematically evaluated as

$T = 6 * F$

$= 6* 1398.6$

$T =8391.6 N$

6 0

3 years ago

A ball is thrown horizontally from the top of a building 0.10 km high. The ball strikes the ground at a point 65 m horizontally

aivan3 [116]

Answer:

option B

Explanation:

given,

height of building = 0.1 km

ball strikes horizontally to ground at = 65 m

speed at which the ball strike = ?

vertical velocity = 0 m/s

time at which the ball strike

$s = \dfrac{1}{2}gt^2$

$t = \sqrt{\dfrac{2s}{g}}$

$t = \sqrt{\dfrac{2\times 100}{9.8}}$

t = 4.53 s

vertical velocity at the time 4.53 s = g × t = 9.8 × 4.53 = 44.39 m/s

horizontal velocity = $\dfrac{65}{4.53}$ =14.35 m/s

speed of the ball = $\sqrt{44.39^2+14.35^2}$

= 46.65 m/s

hence, the speed of the ball just before it strike the ground = 47 m/s

The correct answer is option B

5 0

2 years ago

0.16 mol of argon gas is admitted to an evacuated 70 cm^3 container at 30°C. The gas then undergoes an isothermal expansion to a

Semmy [17]

Answer:

The final pressure of the gas is 9.94 atm.

Explanation:

Given that,

Weight of argon = 0.16 mol

Initial volume = 70 cm³

Angle = 30°C

Final volume = 400 cm³

We need to calculate the initial pressure of gas

Using equation of ideal gas

$PV=nRT$

$P_{i}=\dfrac{nRT}{V}$

Where, P = pressure

R = gas constant

T = temperature

Put the value in the equation

$P_{i}=\dfrac{0.16\times8.314\times(30+273)}{70\times10^{-6}}$

$P_{i}=5.75\times10^{6}\ Pa$

$P_{i}=56.827\ atm$

We need to calculate the final temperature

Using relation pressure and volume

$P_{2}=\dfrac{P_{1}V_{1}}{V_{2}}$

$P_{2}=\dfrac{56.827\times70}{400}$

$P_{2}=9.94\ atm$

Hence, The final pressure of the gas is 9.94 atm.

3 0

2 years ago

Can someone help me plzzz..<br>whoever answers the best will be marked as brainliest.....

IrinaK [193]

Answer:

1) 3 applications of pressure in daily life are :-

● The area of sharp edge of knife, scissor or handsaws are much less then blunt edge. So, for same total force pressure is more for sharp edges than the blunt one. Hence sharp knife, scissors etc, cuts easily than a blunt one.

●Broad handles in bags and suitcases are provided for the comfort. Broad handles have large area. So, the pressure exerted on hands and shoulders would be small while carrying the bags and the suitcases.

●Trucks carrying heavy loads have more than four tyres. More tyres in case of trucks increase the area of contact with the road. This results in reduced pressure on the tyres.

2) Area of the surface which is on ground = 1.5×1

= 1.5m^2

Mass of the block = 300kg

Force applied by the block = Mass × g = 300×10

= 3000N (where g = acceleration due to gravity )

Pressure = Force applied / Area of the surface

= 3000N / 1.5m^2

= 2000 Pa

a) The above experiment signifies that more the area of the surface of an object , less the pressure an object applies.

b) B exerts the minimum pressure because the area of its surface to ground is greater than others & as it has more area of surface , it exerts less pressure. ( area is inversely proportional to pressure )

c) D exerts the maximum pressure because the area of its surface to ground is lesser than others & as it has less area of surface , it exerts more pressure. ( area is inversely proportional to pressure )

d) It depend upon the way an object is kept on ground. If an object is kept in such a way dat the area of the surface to the ground is more , then pressure will be least exerted .If an object is kept in such a way dat the area of the surface to the ground is less, then pressure will be exerted more .

e) Do it yourself . only i will suggest that make the tip of the cone ( which is to the ground ) more narrower.

6 0

3 years ago