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Nastasia [14]
3 years ago
9

What is force and matter plz answer ​

Physics
2 answers:
Travka [436]3 years ago
6 0

Explanation:

<u>FORCE:</u><u> </u>

force is any interaction that, when unopposed, will change the motion of an object.

<u>MATTER:</u>

anything which have mass and occupies space in air is known as Matter

makkiz [27]3 years ago
4 0

Hello There!

A force is simply just a pull or a push on an object. It’s measured in Newtons “N”

Matter is anything that has mass and it takes up space. This is also known as volume

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An ideal gas is enclosed in a piston, and 1600 J of work is done on the gas. As this happens, the internal energy of the gas inc
Phantasy [73]

Answer:

- 1100 J heat flows out

Explanation:

dW = - 1600 J (as work is done on the gas)

dU = 500 J

dQ = ?

According to the first law of thermodynamics

dQ = dU + dW

dQ = 500 - 1600

dQ = - 1100 J

As heat is negative so it flows out.

3 0
3 years ago
What are 3 facts you learned about the periodic table:
d1i1m1o1n [39]

Answer: the rarest element is Francium. J is not on the periodic table. also Dmitri Mendeleev proposed the periodic table.

Explanation: Kinda looked the last one up.

7 0
3 years ago
In an emergency situation, firemen need to respond as quickly as possible. If a fireman is responding from the second floor, how
rjkz [21]

Answer:

4m/s

Explanation:

May be different considering how long the pole is and how heavy the firefighter is.

8 0
3 years ago
A 0.500-kg potato is fired at an angle of 80.0° above the horizontal from a PVC pipe used as a "potato gun" and reaches a height
lions [1.4K]

Answer:

(a) 47.15ms^{-1}

(b) 2470.13ms^{-2}

(c) 1235.06N and 252.05 as a ratio

Explanation:

From Newton's second law of motion

F=ma where m is mass, a is acceleration and F is net force

Also from kinetic equation of motion, velocity and displacement are related using equation

v^{2}=u^{2}+2sa

Where v is final velocity, u is initial velocity, a is acceleration and s is displacement

From the free body diagram attached, final velocity at maximum height is 0 and initial velocity is usin80^{o}

Also, the vertical component can be written as

v_{y}^{2} }=u_{y}^{2} } -2gs The negative sign before 2gs means displacement is opposite the gravitational force

Where g is acceleration due to gravity,u_{y} is vertical component of initial velocity and v_{y} is vertical component of final velocity

Since v_{y} is 0

u_{y}^{2} } =2gs

s=110 and g is taken as 9.8

u_{y}=\sqrt{2*9.8*110}=46.43275

u_{y}=46.43ms^{-1}

Also, it's evident that the vertical component of initial velocity is u_{y}=u_{i}sin \theta where \theta is angle of projection and u_{i} is resultant velocity

Making u_{i} the subject we obtain u_{i}=\frac {u_{y}}{sin \theta}

Since u_{y} and \theta are known as 46.43ms^{-1} and 80^{o} respectively, then u_{i}=\frac {46.43ms^{-1}}{sin 80^{o}}=47.15ms^{-1}

Therefore, the velocity of potato is 47.15ms^{-1}

(b)

Displacement depends on length of tube hence s=0.450m hence going back to kinetic equation v^{2}=u^{2}+2sa

The final velocity v is answer in part a which is 47.15ms^{-1}, initial velocity u is 0ms^{-1} hence the equation is re-written as

v^{2}=2sa and making a the subject we obtain

a=\frac {v^{2}}{2s}

a=\frac {47.15^{2}}{2*0.450}=2470.13ms^{-2}

Therefore, average acceleration is 2470.13ms^{-2}

(c)

From Newton's second law of motion, F=ma where m=0.500kg and a is 2470.13ms^{-2}

Therefore, the average force of potato is

F=0.5*2470.13=1235.06N

F=1235.06N

The weight, W of potato is given by W=mg

Taking R as ratio of average force and weight of potato

R=\frac {F}{W}=\frac {F}{mg} and since F=1235.06, m=0.500kg and g=9.8

R=\frac {1235.06}{0.500*9.8}=252.05

Therefore, ratio of average force to weight is 252.05

8 0
4 years ago
Bro why cant I post this T-T
Leni [432]

Answer: 12.0 m/s^2

Explanation:

Let \alpha be the angular acceleration of the end of the rod

Taking torque about the link, we have:

\tau = W \times OM\\ or\\ \tau = mg\times \left(\dfrac{L}{2}\right)\sin 55^\circ ....(i)

Torque is also given in terms of moment of inertia of the rod and its angular acceleration i.e.

\tau = I_{rod}\ \alpha......(ii)

From equations (i) and (ii) we have:

mg\times \left(\frac{L}{2}\right)\sin 55^\circ = \left(\frac{mL^2}{3}\right)\alpha\ \ \ \ \ \ \ (\because I_{rod} = \dfrac{mL^2}{3}) \\ \\ \alpha = \left(\frac{3g}{2L}\right)\sin 55^\circ\\ \\ \alpha = \left(\frac{3\times 9.8}{2\times 2.4}\right)\sin 55^\circ\\ \\ \boxed{\alpha = 5.02\ rad/s^2} }

The acceleration of the end of the rod farthest from the link is given by:

a = L\alpha\\ \\ a = (2.4\ m)(5.02\ rad/s^2)\\ \\ \boxed{a = 12.0\ m/s^2}

7 0
2 years ago
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