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kykrilka [37]
2 years ago
11

Jake uses a fire extinguisher to put out a small fire. When he squeezes the handle, the flame rettardant is released from the ex

tinguisher with considerable force. As a result, Jake has to reinforce his grip on the extinguisher to hold it in place as it pushes back into his hands. The projectile in this scenario is the . The recoil in this scenario is the . The force on the flame rettardant is the force on the extinguisher. The mass of the flame rettardant is the mass of the extinguisher. The acceleration of the flame rettardant is the acceleration of the extinguisher.
Physics
2 answers:
Tpy6a [65]2 years ago
7 0
Can you attach a picture of the actual problem?
pogonyaev2 years ago
5 0

The projectile in this scenario is the flame r.

The recoil in this scenario is the extinguisher.

The force on the flame r is equal to the force on the extinguisher.

The mass of the flame r is less than the mass of the extinguisher.

The acceleration of the flame r is greater than the acceleration of the extinguisher.

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Which lanyard provides an impact force of less than 1,800 pounds, as recommended by good practices?
enyata [817]

Answer:

Energy absorbing lanyard as per OSHA

Explanation:

Energy absorbing lanyard if working over 6 feet in height so you don't break your back when you fall.

4 0
2 years ago
A sphere of radius 5.00 cm carries charge 3.00 nC. Calculate the electric-field magnitude at a distance 4.00 cm from the center
OlgaM077 [116]

Answer:

a)   E = 8.63 10³ N /C,  E = 7.49 10³ N/C

b)   E= 0 N/C,  E = 7.49 10³ N/C  

Explanation:

a)  For this exercise we can use Gauss's law

         Ф = ∫ E. dA = q_{int} /ε₀

We must take a Gaussian surface in a spherical shape. In this way the line of the electric field and the radi of the sphere are parallel by which the scalar product is reduced to the algebraic product

The area of ​​a sphere is

        A = 4π r²

 

if we use the concept of density

        ρ = q_{int} / V

        q_{int} = ρ V

the volume of the sphere is

      V = 4/3 π r³

         

we substitute

         E 4π r² = ρ (4/3 π r³) /ε₀

         E = ρ r / 3ε₀

the density is

         ρ = Q / V

         V = 4/3 π a³

         E = Q 3 / (4π a³) r / 3ε₀

         k = 1 / 4π ε₀

         E = k Q r / a³

 

let's calculate

for r = 4.00cm = 0.04m

        E = 8.99 10⁹ 3.00 10⁻⁹ 0.04 / 0.05³

        E = 8.63 10³ N / c

for r = 6.00 cm

in this case the gaussine surface is outside the sphere, so all the charge is inside

         E (4π r²) = Q /ε₀

         E = k q / r²

let's calculate

         E = 8.99 10⁹ 3 10⁻⁹ / 0.06²

          E = 7.49 10³ N/C

b) We repeat in calculation for a conducting sphere.

For r = 4 cm

In this case, all the charge eta on the surface of the sphere, due to the mutual repulsion between the mobile charges, so since there is no charge inside the Gaussian surface, therefore the field is zero.

         E = 0

In the case of r = 0.06 m, in this case, all the load is inside the Gaussian surface, therefore the field is

        E = k q / r²

      E = 7.49 10³ N / C

6 0
2 years ago
A certain satellite travels in an approximately circular orbit of radius 2.0 × 106 m with a period of 7 h 11 min. Calculate the
kap26 [50]

Answer: Mass of the planet, M= 8.53 x 10^8kg

Explanation:

Given Radius = 2.0 x 106m

Period T = 7h 11m

Using the third law of kepler's equation which states that the square of the orbital period of any planet is proportional to the cube of the semi-major axis of its orbit.

This is represented by the equation

T^2 = ( 4π^2/GM) R^3

Where T is the period in seconds

T = (7h x 60m + 11m)(60 sec)

= 25860 sec

G represents the gravitational constant

= 6.6 x 10^-11 N.m^2/kg^2 and M is the mass of the planet

Making M the subject of the formula,

M = (4π^2/G)*R^3/T^2

M = (4π^2/ 6.6 x10^-11)*(2×106m)^3(25860s)^2

Therefore Mass of the planet, M= 8.53 x 10^8kg

5 0
1 year ago
A sports car accelerates from 0 to 30 mph in 1.5 s. How long would it take to accelerate from 0 to 60 mph, assuming the power of
Crank

Answer:

6 s

Explanation:

given,

Sports car accelerate from 0 to 30 mph in 1.5 s

time taken to accelerate  0 to 60 mph = ?

The power of the engine is independent of velocity and neglecting friction

power =

P = constant  

the kinetic energy for 60 mph larger than this of 30 mph

 = \dfrac{\dfrac{1}{2}mv_1^2}{\dfrac{1}{2}mv_2^2}

 = \dfrac{v_1^2}{v_2^2}

 = \dfrac{60^2}{30^2}

 = 4

gain in kinetic energy  = P x t

time = 4 x 1.5

       = 6 s

8 0
2 years ago
A uniform cube with mass 0.700 kg and volume 0.0270 m3 is sitting on the floor. A uniform sphere with radius 0.400 m and mass 0.
Sav [38]

Answer:

  44 1/3 cm

Explanation:

The cube has an edge length of ∛0.027 m = 0.3 m, so a center of mass (CoM) 15 cm above the floor.

The sphere's center of mass is 40 cm above the top of the cube, so is 70 cm above the floor. The weighted average of the CoM locations is ...

  ((15 cm)(0.700 kg) +(70 cm)(0.800 kg))/(0.700 kg +0.800 kg)

  = (10.5 kg·cm +56 kg·cm)/(1.500 kg) = 44.333... cm

The center of mass of the two-object system is 44 1/3 cm above the floor.

_____

<em>Comment on the units</em>

We're not familiar with "hcm" as a unit. We presume that you can convert the given answer to the units you desire.

6 0
2 years ago
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