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3 years ago

would the density of a person be the same on the surface of the earth and on the surface of the moon?

Physics

1 answer:

Zielflug [23.3K]3 years ago

7 0

It would remain same. Density does not change with physical environment!!

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I need help with question 2 and 2 please

vitfil [10]

Alright here the answer to number 2

5 0

3 years ago

n an object. one force is 3N to the east and the other force is 9n to the west. what is the net force acting on the object

cricket20 [7]

Answer:

-6N

Explanation:

The force to the east is acting in the positive x-direction therefore it is positive. The force to the east is in the negative x-direction therefore it is negative. The net force is just the sum of the two so 3-9=-6

4 0

2 years ago

Plane a travels at 900km/h and plane b travels at 250/5.which plane travels faster

vesna_86 [32]

Explanation:

We have,

Speed of plane a is 900 km/h

Plane b is moving at a rate of $\dfrac{250\ km}{5\ h}=50\ km/h$

It is required to find which plane is faster. To find which plane is faster, we need to compare their speeds.

Speed of a plane a is 900 km/h and that of plane b is 50 km/h. So, we can say that plane a is moving faster.

5 0

3 years ago

An 8.5 kg crate is pulled 5.1 m up a 30 degree incline by a rope angled 17 degrees above the incline. The tension in the rope is

Ulleksa [173]

Answer:

1. a $W_t=746.63 J$

b $E_p=212.415 J$

c $W_n=183.96J$

2. $T_e=99.71J$

Explanation:

a). The work done by the tension is:

$W_t=T*dt$

$dt=\frac{5.1}{cos(17)}$

$W_t=140N*\frac{5.1}{cos(17)}$

$W_t=746.63 J$

b). The work done potential of gravity

$E_p=m*g*h$

$h=5.1*sin(17)$

$E_p=8.5kh*9.8*5.1*sin(30)$

$E_p=212.415 J$

c). The work done by the normal force

$W_n=N*d_n$

$d_n=5.1*sin(30)=2.55$

$W_n=8.5kg*9.8*cos(30)*2.55$

$W_n=183.96J$

2. The increase in thermal energy is:

$T_e=F*d$

$F_k=u_k*m*g=0.271*8.5kg*9.8*cos(30)$

$F_k=19.5N$

$T_e=19.55*5.1m$

$T_e=99.71J$

4 0

3 years ago

The strength of the electric field at a certain distance from a point charge is represented by E. What is the strength of the el

Nana76 [90]

Answer:

Explanation:

Using Coulomb's law equation

Force of the charge = k qQ /d²

and E = F/ q

substitute for F

E = ( K Qq/ d² ) / q

q cancel q

E = KQ / d²

so twice the distance of the from the point charge will lead to the E ( electric field ) decrease by a 4 = E/4. E is inversely proportional to d²

7 0

3 years ago