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Len [333]
3 years ago
14

A fire hose is made of six sections, each 15 feet in length. The sections are joined together with Storz couplings, which are fo

ur inches long each. What is the total length of the hose?
Physics
1 answer:
gayaneshka [121]3 years ago
8 0

Answer:

h=740\ in

Explanation:

Given:

length of each pipe section, l=15\ ft=180\ in

length of each coupling, c=4 \ in

Now according to question we have six hoses which will require 5 couplings to join them.

So the total length of the hose:

h=6\times l+5\times c

h=6\times 180+5\times 4

h=740\ in is  the total length of the hose.

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A proud new Jaguar owner drives her car at a speed of 25 m/s into a corner. The coefficients of friction between the road and th
ehidna [41]

Answer:

ac = 3.92 m/s²

Explanation:

In this case the frictional force must balance the centripetal force for the car not to skid. Therefore,

Frictional Force = Centripetal Force

where,

Frictional Force = μ(Normal Force) = μ(weight) = μmg

Centripetal Force = (m)(ac)

Therefore,

μmg = (m)(ac)

ac = μg

where,

ac = magnitude of centripetal acceleration of car = ?

μ = coefficient of friction of tires (kinetic) = 0.4

g = 9.8 m/s²

Therefore,

ac = (0.4)(9.8 m/s²)

<u>ac = 3.92 m/s²</u>

5 0
2 years ago
Name the type of component that has a greater resistance as the current through it increases​
Gnesinka [82]

Answer:

filament bulb, filament lamp

Explanation:

4 0
3 years ago
A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.25 kg to a friend standing in front of him
juin [17]

Answer:

The velocity of the student has after throwing the book is 0.0345 m/s.

Explanation:

Given that,

Mass of book =1.25 kg

Combined mass = 112 kg

Velocity of book = 3.61 m/s

Angle = 31°

We need to calculate the magnitude of the velocity of the student has after throwing the book

Using conservation of momentum along horizontal  direction

m_{b}v_{b}\cos\theta= m_{c}v_{c}

v_{s}=\dfrac{m_{b}v_{b}\cos\theta}{m_{c}}

Put the value into the formula

v_{c}=\dfrac{1.25\times3.61\times\cos31}{112}

v_{c}=0.0345\ m/s

Hence, The velocity of the student has after throwing the book is 0.0345 m/s.

3 0
3 years ago
Given the following situation of marble in motion on rolling 10 m/s horizontally from a height of 1.5-m with negligible friction
Norma-Jean [14]

Answer:

The ball would hit the floor approximately 0.55\; \rm s after leaving the table.

The ball would travel approximately 5.5\; \rm m horizontally after leaving the table.

(Assumption: g = 9.81\; \rm m \cdot s^{-2}.)

Explanation:

Let \Delta h denote the change to the height of the ball. Let t denote the time (in seconds) it took for the ball to hit the floor after leaving the table. Let v_0(\text{vertical}) denote the initial vertical velocity of this ball.

If the air resistance on this ball is indeed negligible:\displaystyle \Delta h = -\frac{1}{2}\, g\, t^{2} + v_0(\text{vertical}) \cdot t.

The ball was initially travelling horizontally. In other words, before leaving the table, the vertical velocity of the ball was v_0(\text{vertical}) = 0 \; \rm m \cdot s^{-1}.

The height of the table was 1.5\; \rm m. Therefore, after hitting the floor, the ball would be 1.5\; \rm m \! below where it was before leaving the table. Hence, \Delta h = -1.5\;\rm m.

The equation becomes:

\displaystyle -1.5 = -\frac{9.81}{2} \, t^{2}.

Solve for t:

\displaystyle t = \sqrt{1.5 \times \frac{2}{9.81}} \approx 0.55.

In other words, it would take approximately 0.55\; \rm s for the ball to hit the floor after leaving the table.

Since the air resistance on the ball is negligible, the horizontal velocity of this ball would be constant (at v(\text{horizontal}) =10\; \rm m \cdot s^{-1}) until the ball hits the floor.

The ball was in the air for approximately t = 0.55\; \rm s and would have travelled approximately v(\text{horizontal})\cdot t \approx 5.5\;\rm m horizontally during the flight.

4 0
2 years ago
A net force of 125 n is applied to a certain object. as a result, the object accelerates with an acceleration of 24.0 m/s2. the
pantera1 [17]
Newton's second law states that Fnet = ma, where Fnet is the net force applied, m is the mass of the object, and a is the object's acceleration. You have the values for Fnet and a, so you simply use this equation to solve for m, mass.
8 0
3 years ago
Read 2 more answers
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