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Zarrin [17]
3 years ago
11

A sled whose total mass with cargo is 30.0 kg rests on ice. The coefficient of static friction is 0.20 and the coefficient of ki

netic friction is 0.15 the sled is attached to a rope horizontally in which the tension force is slowly increased how much tension force applied by the rope will cause the sled to start moving?
A. 0 N
B. 9.8 N
C. 59 N
D. 290 N
Physics
1 answer:
romanna [79]3 years ago
5 0

when sled just start to move the force on the sled will be equal to the static friction

So here we need to find the value of static friction

We know that

m = 30 kg

\mu_s = 0.20

now we know that normal force on the block is counterbalanced by weight of the block

F_n = mg

F_n = 30*9.8

now in order to find the friction force we can use

F_s = \mu_s * F_n

F_s = 0.20 * 30 * 9.8

F_s = 59 N

so it requires 59 N of force to move the sled

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A water pump is a positive displacement-type pump true or false
Kay [80]

Answer: True

A water pump belong to a positive displacement pump that provides constant flow of water at fixed speed, regardless of changes in the counter pressure. The two main types of positive displacement pump are rotary pumps and reciprocating pumps.

Moreover, water pump is a reciprocating positive displacement pump that have an expanding cavity on the suction side and a decreasing cavity on the discharge side. In water pumps, the liquid flows into the pumps as the cavity on the suction side expands and then the liquid flows out of the discharge as the cavity collapses providing water in a pail.

6 0
3 years ago
a 2100-kg car drives with a speed of 18 m/s onb a flat road around a curve that has a radius of curvature of 83m. The coefficien
Law Incorporation [45]

Answer:

<u>The magnitude of the friction force is 8197.60 N</u>

Explanation:

Using the definition of the centripetal force we have:

\Sigma F=ma_{c}=m\frac{v^{2}}{R}

Where:

  • m is the mass of the car
  • v is the speed
  • R is the radius of the curvature

Now, the force acting in the motion is just the friction force, so we have:

F_{f}=m\frac{v^{2}}{R}

F_{f}=2100\frac{18^{2}}{83}

F_{f}=8197.60 \: N

<u>Therefore the magnitude of the friction force is 8197.60 N</u>

I hope it helps you!

7 0
3 years ago
Find the lengths of each of the following vectors
Irina18 [472]

Answer:

Explanation:

Generally, length of vector means the magnitude of the vector.

So, given a vector

R = a•i + b•j + c•k

Then, it magnitude can be caused using

|R|= √(a²+b²+c²)

So, applying this to each of the vector given.

(a) 2i + 4j + 3k

The length is

L = √(2²+4²+3²)

L = √(4+16+9)

L = √29

L = 5.385 unit

(b) 5i − 2j + k

Note that k means 1k

The length is

L = √(5²+(-2)²+1²)

Note that, -×- = +

L = √(25+4+1)

L = √30

L = 5.477 unit

(c) 2i − k

Note that, since there is no component j implies that j component is 0

L = 2i + 0j - 1k

The length is

L = √(2²+0²+(-1)²)

L = √(4+0+1)

L = √5

L = 2.236 unit

(d) 5i

Same as above no is j-component and k-component

L = 5i + 0j + 0k

The length is

L = √(5²+0²+0²)

L = √(25+0+0)

L = √25

L = 5 unit

(e) 3i − 2j − k

The length is

L = √(3²+(-2)²+(-1)²)

L = √(9+4+1)

L = √14

L = 3.742 unit

(f) i + j + k

The length is

L = √(1²+1²+1²)

L = √(1+1+1)

L = √3

L = 1.7321 unit

3 0
3 years ago
A car with a mass of 1600 kg is towing a trailer with a mass of 420 kg. The car
Gekata [30.6K]

Answer:

1.63366

Explanation:

I got this answer from calculator soups physics calculators. I really recommend their website for formulas.  

3 0
3 years ago
Convierta 164 decimetros a hectometros
-BARSIC- [3]

Answer:

sinco

Explanation:

6 0
3 years ago
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