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zaharov [31]
3 years ago
7

all objects near the earths surface-regardless of size and weight have the same force of gravity acting on them. is it true or f

alse?
Physics
1 answer:
pickupchik [31]3 years ago
5 0

False: the force of gravity acting on different objects is different and depends on their mass

Explanation:

The answer is false.

The force of gravity acting on an object (also known as weight) near the Earth's surface is given by:

F=mg

where:

m is the mass of the object

g=9.8 m/s^2 is the acceleration of gravity

We see from the formula that the force of gravity acting on an object depends on the mass: the larger the mass of the object, the stronger the gravitational force acting on it, and the smaller the mass, the weaker the force of gravity.

The factor that does not change is the acceleration of gravity, which is constant (9.8 m/s^2) if we are near the Earth's surface, and implies that all the objects in free fall accelerate at the same rate towards the ground, regardless of their size and weight.

Learn more about forces and weight here:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

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Two ropes are attached to a 35 kg object. The first rope applies a force of 20 N and the second applies a force of 55 N. If the
Goshia [24]

Answer:

a=1.672\ m.s^{-2}

Explanation:

Given:

  • mass of the object, m=35\ kg

forces by two mutually perpendicular ropes of the attached to the object:

  • F_x=20\ N
  • F_y=55\ N

<u>Now we find the resultant force effect due to the two given forces:</u>

F=\sqrt{F_x^2+F_y^2}

F=\sqrt{(20)^2+(55)^2}

F=58.52\ N

<u>Now the acceleration will be due to this resultant force:</u>

a=\frac{F}{m}

a=\frac{58.52}{35}

a=1.672\ m.s^{-2}

3 0
3 years ago
Read 2 more answers
A truck travels down the highway at a speed of 110 km/hr. How long does the trip last if the truck covered 2200km?
vivado [14]
It'll last 20 hours. If it travels 110 miles in one hours, 110 times 20 equals 2,200.
8 0
3 years ago
High-energy particles are observed in laboratories by photographing the tracks they leave in certain detectors; the length of th
murzikaleks [220]

Answer:

Lifetime = 4.928 x 10^-32 s  

Explanation:

(1 / v2 – 1 / c2) x2 = T2

T2 = (1/ 297900000 – 1 / 90000000000000000) 0.0000013225

T2 = (3.357 x 10^-9 x 1.11 x 10^-17) 1.3225 x 10^-6

T2 = (3.726 x 10^-26) 1.3225 x 10^-6 = 4.928 x 10^-32 s  

4 0
3 years ago
Consider the following FBDs.FBDsAn elevator is suspended by a cable and moves uniformly upward. Which of the above free body dia
Drupady [299]

The force of gravity F_g will act downwards.

Normal force F_N will act upwards equal to the force of gravity.

A force due to uniform acceleration F_a will act upwards to move the elevator upwards.

Thus, figure E is the correct answer.

6 0
1 year ago
A 150 g copper bowl contains 210 g of water, both at 24.0°C. A very hot 430 g copper cylinder is dropped into the water, causing
Dahasolnce [82]

Answer:

A. 15969.22 cal

B. 1052,22 cal

C. 528,87 °C

Explanation:

To solve this kind of question, a proper method is to work from the data that you have towards the data that you need. Also, it is recommended to analyze related equations as they could give us clues on how to find the missing information or the information that the problem is asking us.

Let us start with Question A. It is important to remember that energy transfers with the environment are being neglected; this means that all the energy that the cylinder lose is picked up by the water and the copper bowl. To find the amount of energy transferred to the water, we first find the amount of energy necessary to raise the water’s temperature to 100°C and then we find the amount of energy necessary to evaporate the 17.1 g of water indicated by the question. This would be:

Q = m_water * CP_water *∆T =210g *1 cal/(g K) * (100°C-24°C) = 15960 cal

Q_evap = m_wat * L = 17,1 g * 539 cal/kg* (1 kg)/(1000 g) =9.2169 cal

Therefore, the total energy that was transferred to the water is the sum of these components, that would be Q_tot = 15960 cal + 9.2159 cal = 15969.22 cal.  Let´s also remember that a temperature difference in K is equal to a temperature difference in ° C

To solve Question B, we use the same method. We must find the amount of energy necessary to raise the temperature from its initial temperature to the one stated by the problem to be the equilibrium temperature of the system (100°C):

Q= m_copper *CP_copper *∆T = 150g * 0.0923 cal/(g K) * (100°C-24°C) = 1052,22 cal

If we add the components we just found in questions A and B, we can find the amount of energy than the Copper cylinder lost, this would be: Q_tot = 15969.22 cal + 1052.22 cal = 17021.44 cal.

The question C asks us to find the initial temperature of the cylinder and Q_tot will help us to find it.

We know that Q_tot is the energy lost by the cylinder and we also know that Q_tot = m_cylinder * CP_copper * ∆T. Therefore, what we need to do  is clear the last term of the equation and find the initial temperature.

Q_tot = m_cylinder *CP_copper *∆T → T_fin-T_initial = Q_tot/(m_cylinder*CP_copper ) = (-17021.44 cal)/(430g*0.0923 cal/(g K))

→ T_initial = 100°C + (-17021.44 cal)/(430g * 0.0923 cal/(g K)) = 528,87 °C

If we convert the 100°C to K before we do the calculation, the result would be the same one, You would only need to add 273,15 to the final result to check it out.  

Hope everything was clear. If you have any further question, I'll be happy to help :D

5 0
3 years ago
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