Do the math for 30%÷75

Two blocks joined by a string have masses of 6 and 9 kg. They rest on a frictionless horizontal surface. A 2nd string, attached

Tom [10]

Answer:

12N

Explanation:

Suppose the string mass is negligible, the total mass of the 2 block system is 6 + 9 = 15 kg

So the acceleration of the system when subjected to 30N force is

a = F / M = 30 / 15 = 2 m/s2

So both blocks would have the same acceleration, however, the force acting on the 6kg block would have a magnitude of

f = am = 2 * 6 = 12N

This is the tension in the string between the blocks

5 0

3 years ago

On a planet where g = 10.0 m/s2 and air resistance is negligible, a sled is at rest on a rough inclined hill rising at 30°. The

Cloud [144]

Answer:

Explanation:

a is the acceleration

μ is the coefficient of friction

Acceleration of the object is given by

$a = g (\sin \theta -\mu \cos \theta)\\\\=10( \sin 30 - 0.4 \cos 30)\\\\=10(0.5-0.3464)\\\\=1.54m/s^2$

Velocity at the bottom

$v^2=u^2+2as\\\\u=0\\\\v^2=2as\\\\=2*1.54*8\\\\=24.576\\\\v=4.96m/s$

after travelling 4m , its velocity becomes 0

$a=\frac{v^2-u^2}{2s}$

$a=\frac{0-u^2}{2s}$

$a=\frac{-(-4.96)^2}{2*4} \\\\=-3.075m/s^2$

Coefficient of kinetic friction

μ = F/N

$=\frac{ma}{mg} \\\\=\frac{3.075}{10} \\\\=0.31$

Therefore, the Coefficient of kinetic friction is 0.31

8 0

3 years ago

Read 2 more answers

A crate with a mass of m = 450 kg rests on the horizontal deck of a ship. The coefficient of static friction between the crate a

Zielflug [23.3K]

Answer: $F_{v} =\mu_{k} mg$

Magnitude of the force is 2601.9 N

Explanation:

m = 450 kg

coefficient of static friction μs = 0.73

coefficient of kinetic friction is μk = 0.59

The force required to start crate moving is $F_{s} =\mu_{s} mg$ .

but once crate starts moving the force of friction is reduced $F_{v} =\mu_{k} mg$ .

Hence to keep crate moving at constant velocity we have to reduce the force pushing crate ie $F_{v} =\mu_{k} mg$ .

Then the above pushing force will equal the frictional force due to kinetic friction and constant velocity is possible as forces are balanced.

Magnitude of the force

$F_{v} =\mu_{k} mg\\F_{v} =0.59 \times 450 \times 9.8\\F_{v} =2601.9 N$

4 0

4 years ago

he Hobbits are building a watchtower so they can prepare to battle in case trolls decide to attack them. One Hobbit will always

joja [24]

Answer:

5m/8

Explanation:

Function T gives the time the Hobbits have to prepare for the attack, T(k), in minutes, as a function of troll's distance, k, in meters.

Function V gives visibility from the watchtower, V(m), in meters, as a function of the height of the watchtower, m, in meters.

Therefore, T(V(m)) will give the time the Hobbits have to prepare for the troll attack as a function of the height, m, of the watchtower.

We can input m into function V to obtain the visibility from watchtower, V(m), in meters. Since visibility indicates the distance you can see, this also gives the distance of the trolls. This can then be input into function T to obtain the time that the Hobbits have to prepare for a troll attack.

Let's find T(V(m)) by substituting the formula for V(m) into function T as shown below.

T(V(M))=T(50m)

=50m/80

We can simplify this as follows:

=50m/80

=5m/8

4 0

3 years ago

A volleyball is dropped from a cliff and a soccer ball is thrown upward from the same position. When each ball reaches the groun

vredina [299]

The question is whether the statement is true or false.

The answer if false.

Explanation:

It is exactly the opposite. The soccer ball will hit the ground with greater velocity.

Since the soccer ball is thrown upward, when it returns to the same heigth from which it was throwm it will have a velocity downward, which will make that the soocer ball reaches the ground at the bottom of the clif with greater velocity than the volleball.

The greater the velocity with which the soccer ball is thrown upward, the greater its velocity when reaches the same point from which it was thrown, and the greater the velocity with which it will hit the ground at the bottom of the clif.

6 0

3 years ago