Which form of friction produces the least amount of force?

After a 50-kg person steps from a boat onto the shore, the boat moves away with a speed of 0.70 m/s with respect to the shore. I

Svetradugi [14.3K]

Answer:

M=125 kg

v=1.75 m/s

Explanation:

From the law of linear momentum

P =mv

Case 1 50*V =M* 0.7 equation 1

50*V =(M+50)* 0.5 equation 2

equating 1 and 2

M* 0.7=(m+50)* 0.5

0.2 M= 25

M=125 kg

Putting value of M in equation 1

50*V =125*0.7

V=1.75 m/s

7 0

3 years ago

Read 2 more answers

A plastic bottle partially filled with water floats on water, even though the density of the plastic (1.2 g/cc) is more than tha

Semmy [17]

Answer:

True the plastic will float because of the principle of flotation or buoyancy

Explanation:

Buoyancy explains it all!!

Buoyancy is the upward force/upthrust experienced by a body immersed totally or partially in a liquid.

According to the principle of flotation:

"when a body is totally or partially immersed in liquid it experiences an upthrust which is equal to the volume of fluid displaced"

The plastic will float due to the fact the average density of the total volume of the plastic and the air inside it is less than the same volume of water it is floating in

5 0

3 years ago

A 500 N force accelerates an object at 20 m s-2. What is its mass?

avanturin [10]

Answer: The mass of the object is 25kg.

The given question deals with Newton's second law of motion and its applications.

Explanation: Given force, F=500N

acceleration, a=20 m/ $s^{2}$

From Newton's 2nd law of motion , we have

F=ma where m=mass of the object

⇒500=m×20

⇒m=500/20=25

∴ Mass of the object is 25 kg .

Reference Link: brainly.com/question/1141170

#SPJ2

8 0

2 years ago

Hank and Harry are two ice skaters whiling away time by playing 'tug of war' between practice sessions. They hold on to opposite

Alex73 [517]

Answer:

the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1

Explanation:

Given the data in the question;

Hank and Harry are two ice skaters, since both are on top of ice, we assume that friction is negligible.

We know that from Newton's Second Law;

Force = mass × Acceleration

F = ma

Since they hold on to opposite ends of the same rope. They have the same magnitude of force |F|, which is the same as the tension in the rope.

Now,

Mass $_{Hank$ × Acceleration $_{Hank$ = Mass $_{Henry$ × Acceleration $_{Henry$

so

Mass $_{Hank$ / Mass $_{Henry$ = Acceleration $_{Henry$ / Acceleration $_{Hank$

given that; magnitude of Hank's acceleration is 1.26 times greater than the magnitude of Harry's acceleration,

Mass $_{Hank$ / Mass $_{Henry$ = 1 / 1.26

Mass $_{Hank$ / Mass $_{Henry$ = 0.7937 or [ 0.7937 : 1 ]

Therefore, the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1 ]

8 0

2 years ago

Tutorial Exercise An unstable atomic nucleus of mass 1.83 10-26 kg initially at rest disintegrates into three particles. One of

kogti [31]

Answer:

A) v3 = -[6.29 × 10^(6)]j^ - [7.06 × 10^(6)]i^

B) K_total = 373.08 × 10^(-15) J

Explanation:

We are given;

Mass of unstable atomic nucleus; M = 1.83 × 10^(-26) kg

Mass of first particle; m1 = 5.03 × 10^(-27) kg

Speed of first particle in y-direction; v1 = (6 × 10^(6) m/s) j^

Mass of second particle; m2 = 8.47 × 10^(-27) kg

Speed of second particle in x - direction; v2 = (4 × 10^(6) m/s) i^

Now, we don't have the mass of the third particle but since we are told the unstable atomic nucleus disintegrates into 3 particles, thus;

M = m1 + m2 + m3

1.83 × 10^(-26) = (5.03 × 10^(-27)) + (8.47 × 10^(-27)) + m3

m3 = (1.83 × 10^(-26)) - (13.5 × 10^(-27))

m3 = 4.8 × 10^(-27) kg

A) Applying law of conservation of momentum, we have;

MV = (m1 × v1) + (m2 × v2) + (m3 × v3)

Now, the unstable atomic nucleus was at rest before disintegration, thus V = 0 m/s.

Thus, we now have;

0 = (m1 × v1) + (m2 × v2) + (m3 × v3)

We want to find the velocity of the third particle v3. Let's make it the subject of the formula;

v3 = [(m1 × v1) + (m2 × v2)]/(-m3)

Plugging in the relevant values, we have;

v3 = [(5.03 × 10^(-27) × 6 × 10^(6))j^ + (8.47 × 10^(-27) × 4 × 10^(6))i^]/(-4.8 × 10^(-27))

v3 = [(30.18 × 10^(-21))j^ + (33.88 × 10^(-21))i^]/(-4.8 × 10^(-27))

v3 = -[6.29 × 10^(6)]j^ - [7.06 × 10^(6)]i^

B) Formula for kinetic energy is;

K = ½mv²

Now,total kinetic energy is;

K_total = K1 + K2 + K3

K1 = ½ × 5.03 × 10^(-27) × (6 × 10^(6))²

K1 = 90.54 × 10^(-15) J

K2 = ½ × 8.47 × 10^(-27) × (4 × 10^(6))²

K2 = 67.76 × 10^(-15)

To find K3, let's first find the magnitude of v3 because it's still in vector form.

Thus;

v3 = √[(-6.29 × 10^(6))² + (-7.06 × 10^(6))²]

v3 = 9.46 × 10^(6) m/s

K3 = ½ × 4.8 × 10^(-27) × (9.46 × 10^(6))²

K3 = 214.78 × 10^(-15) J

K_total = (90.54 × 10^(-15)) + (67.76 × 10^(-15)) + (214.78 × 10^(-15))

K_total = 373.08 × 10^(-15) J

7 0

3 years ago