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3 years ago

REALLY NEED HELP ON THESE QUESTIONS WILL MARK BRAINLYEST 100POINTS UP FOR GRABS!!!!!

Physics

2 answers:

Lunna [17]3 years ago

4 0

when an object is pushed by applied force it accelerated by

a = 5.9 m/s^2

now here we will have

$F_{ap} = 38 N$

m = 4.5 kg

$a = 5.9 m/s^2$

now we can say that net force on the object is applied force and friction force

this net force is given by Newton's II law as

$F_{net} = ma$

$F_{ap} - F_f = ma$

$38 - F_f = 4.5*5.9$

$38 - F_f = 26.55$

$F_f = 11.45 N$

so friction force will be 11.45 N

when an object is pushed by applied force it accelerated by

a = 6.5 m/s^2

now here we will have

$F_{f} = 5.6 N$

m = 1.9 kg

$a = 6.5 m/s^2$

now we can say that net force on the object is applied force and friction force

this net force is given by Newton's II law as

$F_{net} = ma$

$F_{ap} - F_f = ma$

$F_{ap} - 5.6 = 1.9*6.5$

$F_{ap} - 5.6= 12.35$

$F_{ap} = 17.95 N$

so applied force will be 17.95 N

Given that

mass of hockey Puck = 0.8 kg

applied force on the puck = 570 N

acceleration of the puck = 682.5 m/s^2

now we can say that net force on the object is applied force and drag force of the air

this net force is given by Newton's II law as

$F_{net} = ma$

$F_{ap} - F_d = ma$

$570 - F_d = 0.8*682.5$

$570 - F_d = 546$

$F_d = 570 - 546 = 24 N$

so drag force due to air on the puck will be 24 N

#7)

Static friction is always stronger than sliding friction because in static situation of the object the bonding of the molecules of two surfaces will be very strong and hence it requires more force to start the motion, once the motion has been started the bonding of two surfaces becomes weak and then object can continue its motion with less friction.

Given that

mass of box = 3.8 kg

applied force on the box = 27 N

acceleration of the box = 4.8 m/s^2

now we can say that net force on the object is applied force and friction force of the ground

this net force is given by Newton's II law as

$F_{net} = ma$

$F_{ap} - F_f = ma$

$27 - F_f = 3.8*4.8$

$27 - F_f = 18.24$

$F_f = 27 - 18.24 = 8.76 N$

so friction force due to ground on the box will be 8.76 N

Ksju [112]3 years ago

3 0

Answer:

8.76 is the answer

Explanation:

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3 years ago

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4 years ago

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When the valve between the 2.00-L bulb, in which the gas pressure is 2.00 atm, and the 3.00-L bulb, in which the gas pressure is

padilas [110]

Answer:

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Explanation:

Let assume that gases inside bulbs behave as an ideal gas and have the same temperature. Then, conditions of gases before and after valve opened are now modelled:

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$P_{A} \cdot V_{A} = n_{A} \cdot R_{u} \cdot T$

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After some algebraic manipulation, a formula for final pressure is derived:

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3 years ago

A projectile was launched horizontally with a velocity of 388 m/s, 2.89 m above the ground. How long did it take the projectile

tamaranim1 [39]

Answer:

Explanation:

Given

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Required

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