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boyakko [2]
2 years ago
13

An iceberg of density 920kg/m^3 floats in seawater of density 1025kg/m^3 with volume of 10^6m^3.What is the total mass of the ic

eberg?
Physics
1 answer:
zhenek [66]2 years ago
5 0
Im pretty sure the answer is 1.95x10^9
You might be interested in
12
LiRa [457]

The horizontal speed of the object 1.0 seconds later is 1) 5.0 m/s.

Explanation:

The motion of an object thrown horizontally off a cliff is a projectile motion, which follows a parabolic path that consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

This means that the horizontal speed of an object in projectile motion does not change, and remains constant during the whole motion.

Since in this case the object has been launched with a horizontal speed of

v = 5.0 m/s

this means that this speed will remain constant during the motion, so its horizontal speed 1.0 s later is also 5.0 m/s.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

7 0
3 years ago
What did newton's law of universal gravitation tell us about how gravity works?
vazorg [7]
The force of gravity form the Sun will be stronger on a n object with more mass
6 0
3 years ago
The cabinet is mounted on coasters and has a mass of 45 kg. The casters are locked to prevent the tires from rotating. The coeff
stira [4]

Answer:

the force P required for impending motion is 132.3 N

the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

Explanation:

Given that:

mass of the cabinet  m = 45 kg

coefficient of static friction μ =  0.30

A free flow body diagram illustrating what the question represents is attached in the file below;

The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.

Thus; considering the forces along the vertical axis ; we have :

\sum f_y =0

The upward force and the downward force is :

N_A+N_B = mg

where;

\mathbf { N_A  \ and  \ N_B} are the normal contact force at center point A and B respectively .

N_A+N_B = 45*9.8

N_A+N_B = 441    ------- equation (1)

Considering the forces on the horizontal axis:

\sum f_x = 0

F_A +F_B  = P

where ;

\mathbf{ F_A \ and \ F_B } are the static friction at center point A and B respectively.

which can be written also as:

\mu_s N_A + \mu_s N_B  = P

\mu_s( N_A +  N_B)  = P

replacing our value from equation (1)

P = 0.30 ( 441)    

P = 132.3 N

Thus; the force P required for impending motion is 132.3 N

b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm

the largest value of "h" allowed for  the cabinet is not to tip over is calculated by determining the limiting condition  of the unbalanced torque whose effect is canceled by the normal reaction at N_A and it is shifted to N_B:  

Then:

\sum M _B = 0

P*h = mg*0.24

h =\frac{45*9.8*0.24}{132.3}

h = 0.8 m

Thus; the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

6 0
3 years ago
A tank initially holds 100 gallons of salt solution in which 50 lbs of salt has been dissolved. A pipe fills the tank with brine
olga_2 [115]

Answer:

A. 171.24 Ibs

Explanation:

To find the amount of salt in the tank,

Let Q = Amount of salt in the mixture

And let 100 + (3-2)t = 100 + t be the volume of mixture at anytime t.

Rate of gain - Rate of loss = dQ / dt

Concentration of salt = Q / (100+t)

For the linear differential equation,

dQ / dt = 3(2) - 2 [Q/ (100 + t)]

dQ /dt + Q [2 / (100 + t)] = 6

The general solution of the linear differential equation is:

Q (i.f) = ∫ A(t) (i.f) dt + C

Therefore,

i.f = e ^ ∫ P(t) dt

And P(t) = 2 / (100 + t)

i.f = e ^ ∫ 2 / (100 + t)

  = e ^ 2㏑ (100 + t)

     = e ^ ㏑ (100 + t) ^2 = (100 + t) ^2

Q(100 + t) ^ 2 = ∫6 (100 + t) ^2 dt + C

 Q(100 + t) ^2 = 2(100 + t) ^ 3 + C

  When t = 0, Q = 50

Therefore,

50( 100) ^2 = 2(100) ^3 + C

 C = -1.5 * 10 ^6

therefore, when t = 30,

Q (100 + 30) ^2 = 2(100 + 30) ^3 - 1.5 * 10 ^6

 Q (400) ^2 = 2(130) ^3 - 1.5 * 10 ^6

    Q = 171.24 Ibs

7 0
3 years ago
A motorboat traveling with a current can go 160 km in 4 hours. against the current it takes 5 hours to go the same distance. Fin
MatroZZZ [7]
<h2>Speed of motorboat is 36 km/hr and speed of current is 4 km/hr.</h2>

Explanation:

Let speed of motor boat be m and speed of current be c.

A motorboat traveling with a current can go 160 km in 4 hours.

   Distance = 160 km

   Time = 4 hours

    Speed = m + c

   We have

            Distance = Speed x Time

            160 = (m+c) x 4

            m + c = 40     --------------------- eqn 1

Against the current it takes 5 hours to go the same distance.

   Distance = 160 km

   Time = 5 hours

    Speed = m - c

   We have

            Distance = Speed x Time

            160 = (m-c) x 5

            m - c = 32     --------------------- eqn 2

eqn 1 + eqn 2

           2m = 40 + 32

             m = 36 km/hr

Substituting in eqn 1

               36 + c = 40

                      c = 4 km/hr

Speed of motorboat is 36 km/hr and speed of current is 4 km/hr.

3 0
3 years ago
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