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2 years ago

In which direction does gravitational force act

Physics

1 answer:

Sunny_sXe [5.5K]2 years ago

4 0

Gravitational force pushes down on us to keep us on the ground. So it pushes in a downward direction.

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A 9.00 g bullet is fired horizontally into a 1.20 kg wooden block resting on a horizontal surface. The coefficient of kinetic fr

Zinaida [17]

Answer:

The initial speed of bullet is "164 m/s".

Explanation:

The given values are:

mass of bullet,

$m'=9.00 \ g$

or,

$=0.009 \ kg$

mass of wooden block,

$m=1.20 \ kg$

speed,

$s=0.390 \ m$

Coefficient of kinetic friction,

$\mu=0.20$

As we know,

The Kinematic equation is:

⇒ $v^2=u^2+2as$

then,

Initial velocity will be:

⇒ $u=v^2-2as$

$=v^2-2 \mu gs$

On substituting the given values, we get

⇒ $u=\sqrt{0-2\times 0.20\times 9.8\times 0.390}$

$=\sqrt{-1.5288}$

$=1.23 \ m/s$

As we know,

The conservation of momentum is:

⇒ $mu=m'u'$

or,

⇒ Initial speed, $u'=\frac{mu}{m'}$

On substituting the values, we get

⇒ $=\frac{1.20\times 1.23}{0.009}$

⇒ $=\frac{1.476}{0.009}$

⇒ $=164 \ m/s$

3 0

3 years ago

Help can’t find the answer no where

katen-ka-za [31]

i'm stuck on that question also

5 0

3 years ago

Radiant heat makes it impossible to stand close to a hot lava flow. Calculate the rate of heat loss by radiation from 1.00 m^2 o

VARVARA [1.3K]

The rate of heat loss by radiation is equal to <u>-207.5kW</u>

Why?

To calculate the heat loss rate (or heat transfer rate) by radiation, from the given situation, we can use the following formula:

$HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )$

Where,

E, is the emissivity of the body.

A, is the area of the body.

T, are the temperatures.

S, is the Stefan-Boltzmann constant, which is equal to:

$5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }$

Now, before substitute the given information, we must remember that the given formula works with absolute temperatures (Kelvin), so, we need to convert the given values of temperature from Celsius degrees to Kelvin.

We know that:

$K=Celsius+273.15$

So, converting we have:

$T_{1}=1110\°C+273.15=1383.15K\\\\T_{2}=36.2\°C+273.15=309.35K$

Therefore, substituting the given information and calculating, we have:

$HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )$

$HeatLossRate=1*5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }*1m^{2} *((309.35K)^{4} -(1383.15})^{4} )\\\\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(95697.42K^{4} -3.66x10^{12}K^{4})\\ \\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(-3.66x10^{12} K^{4})=-207522W=-207.5kW$

Hence, we have that the rate of heat loss is equal to -207.5kW.

8 0

3 years ago

Marking branliest!

Ivenika [448]

Answer:

D)evaluating a solution

Explanation:

In this scenario, the next logical step would be evaluating a solution. This is because Jasper and Samantha have already identified the problem/need which is that the robot needs to be able to move a 10-gram weight at least 2 meters and turn in a circle. They also designed and implemented a solution because they have already built the robot. Therefore the only step missing is to evaluate and make sure that the robot they built is able to complete the requirements.

7 0

3 years ago

A 0.500-kg block slides up a plane inclined at a 30° angle. If it slides 1.50 m before coming to rest while encountering a frict

julia-pushkina [17]

B it’s Intail velocity

7 0

3 years ago