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Elena L [17]
3 years ago
15

Can someone define 'work' for me please? I looked it up but there are a lot of different answers.

Physics
2 answers:
vekshin13 years ago
7 0

Answer:

Work- Activity involving mental or physical effort done in order to achieve a purpose or result.

Explanation:

This was the first definition from the dictionary on google :)

scZoUnD [109]3 years ago
6 0
Be engaged in physical or mental activity in order to achieve a result; do work. a task or tasks to be undertaken; something a person or thing has to do.
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Calculate the force of gravity on the 0.60-kg mass if it were 1.3×107 m above Earth's surface (that is, if it were three Earth r
KIM [24]
The force of gravity between two objects is given by:
F=G \frac{m_1 m_2}{r^2}
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation

In this problem, the mass of the object is m_1=0.60 kg, while the Earth's mass is m_2=5.97 \cdot 10^{24} kg. Their separation is r=1.3 \cdot 10^7 m, therefore the gravitational force exerted on the object is
F=(6.67 \cdot 10^{-11}m^3 kg^{-1} s^{-2}) \frac{(0.60 kg)(5.97 \cdot 10^{24} kg)}{(1.3 \cdot 10^7 m)^2}=1.4 N
5 0
3 years ago
You slide a hockey puck across the ice in a hockey rink
REY [17]

Answer-

mk thats kool.

8 0
3 years ago
Read 2 more answers
If the total momentum of a system is changing:
DENIUS [597]

Answer:

(d) a net external force must be acting on the system

Explanation:

Momentum is given as the product of mass and velocity.

P = MV

According to Newton's second law of motion, " Force applied to a body (system) is directly proportional to the rate of change of momentum of the body (system) which takes place in the direction of the applied force (external force).

F ∝ΔMV

Therefore, If the total momentum of a system is changing, a net external force must be acting on the system.

(d) a net external force must be acting on the system

3 0
3 years ago
A horizontal uniform bar of mass 2.7 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to t
Jlenok [28]

Answer:

14.36 N

Explanation:

T_{1} = Tension in string 1

T_{2} = Tension in string 2

m_b = mass of the bar = 2.7 kg

W_b = weight of the bar

weight of the bar is given as

W_b = m_{b} g = (2.7) (9.8) = 26.46N

m_m = mass of the bar = 1.35 kg

W_m = weight of the monkey

weight of the monkey is given as

W_m = m_{m} g = (1.35) (9.8) = 13.23N

Using equilibrium of torque about left end

W_{m} (AB) + W_{b} (AB) = T_{2} (AC)\\W_{m} (AB) + W_{b} (AB) = T_{2} (AD - CD)\\(13.23) (1.5) + (26.46)(1.5) = T_{2} (3 - 0.65)\\\\T_{2} = 25.33 N

Using equilibrium of force in vertical direction

T_{1} + T_{2} = W_{b} + W_{m}\\T_{1} + 25.33 = 26.46 + 13.23\\T_{1} = 14.36 N

7 0
3 years ago
A solution containing 10.0 g of an unknown liquid and 90.0 g water has a freezing point of -3.33 °C. Given Kf = 1.86 °C/m for wa
Fantom [35]

Answer:

62.06 g/mol

Explanation:

We are given that a solution containing 10 g of an unknown liquid and 90 g

Given mass of solute =W_B=10 g

Given mass of solvent=W_A=90 g

k_f=1.86^{\circ}C/m

Freezing point of solution =-3.33^{\circ}C

Freezing point of solvent =0^{\circ}C

Change in freezing point =Depression in freezing point

=Freezing point of solvent - freezing point of solution=0+3.33=3.33^{\circ}

\Delta T_f=\frac{W_B\times K_f\times 1000}{W_A\times M_B}

M_B=\frac{10\times 1.86\times 1000}{3.33\times 90}

M_B=62.06 g/mol

Hence, molar mass of unknown liquid is 62.06g/mol.

6 0
3 years ago
Read 2 more answers
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