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ki77a [65]
3 years ago
7

Kenna has a mass of 30 kilograms. What's her weight in newtons? Assume that acceleration due to gravity is 9.8 N/kg.

Physics
1 answer:
astra-53 [7]3 years ago
5 0

Answer:

294 N

Explanation:

The weight of an object is the gravitational force exerted by the Earth on the object.

The direction of this force is always towards the Earth's centre, while the magnitude is given by:

W=mg

where

W is the weight

m is the mass of the object

g is the acceleration due to gravity

Here we have:

m = 30 kg is the mass of the object

g = 9.8 N/kg is the acceleration due to gravity

So, the weight of the object is

W=(30)(9.8)=294 N

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A hot-air balloon is ascending at the rate of 10 m/s and is 74 m above the ground when a package is dropped over the side. (a) H
Reika [66]

Answer:

The answer to your question is:

a)  t1 = 2.99 s ≈ 3 s

b)  vf = 39.43 m/s

Explanation:

Data

vo = 10 m/s

h = 74 m

g = 9.81 m/s

t = ?   time to reach the ground

vf = ?   final speed

a)    h = vot + (1/2)gt²

     74 = 10t + (1/2)9.81t²

     4.9t² + 10t -74 = 0                  solve by using quadratic formula

   

   t = (-b ± √ (b² -4ac) / 2a

   t = (-10 ± √ (10² -4(4.9(-74) / 2(4.9)

   t = (-10 ± √ 1550.4 ) / 9.81

  t1 = (-10 + √ 1550.4 ) / 9.81               t2 = (-10 - √ 1550.4 ) / 9.81

  t1 = (-10 ± 39.38 ) / 9.81                    t2 = (-10 - 39.38) / 9.81

   t1 = 2.99 s ≈ 3 s                               t2 = is negative then is wrong there are

                                                                   no negative times.

b) Formula vf = vo + gt

                  vf = 10 + (9.81)(3)

                  vf = 10 + 29.43

                  vf = 39.43 m/s

4 0
2 years ago
A toroid having a square cross section, 5.00 cm on a side, and an inner radius of 15.0 cm has 500 turns and carries a current of
SCORPION-xisa [38]

Answer:

a).β=0.53x10^{-3} T

a).β=0.40 x10^{-4} T

Explanation:

The magnetic field at distance 'r' from the center of toroid is given by:

\beta =\frac{u_{o}*I*N}{2\pi*r}

a).

N=500\\I=0.800A\\r=15cm*\frac{1m}{100cm}=0.15m\\u_{o}=4\pi x10^{-7}\frac{T*m}{A}  \\\beta=\frac{4\pi x10^{-7}\frac{T*m}{A}*0.8A*500}{2\pi*0.15m} \\\beta=0.53x10^{-3}T

b).

The distance is the radius add the cross section so:

r_{1}=15cm+5cm\\r_{1}=20cm

r_{1} =20cm*\frac{1m}{100cm}=0.20m

\beta =\frac{u_{o}*I*N}{2\pi*r1}

\beta =\frac{4\pi x10^{-7}*0.80A*500 }{2\pi*0.20m} \\\beta=0.4x10^{-3} T

3 0
3 years ago
A motorcyclist accelerates from rest to 10 mi/hr. what is the change in velocity
gulaghasi [49]

The change in velocity is 10 mi/h (4.47 m/s)

Explanation:

The change in velocity of the motorcyclist is given by

\Delta v = v-u

where

v is the final velocity

u is the initial velocity

In this problem, we have

u = 0 (the motorbike starts from rest)

v = 10 mi/h

Therefore, the change in velocity is

\Delta v = 10 -0 = 10 mi/h

And keeping in mind that

1 mile = 1609 m

1 h = 3600 s

We can convert it into m/s:

\Delta v = 10 \frac{mi}{h} \cdot \frac{1609 m/mi}{3600 s/h}=4.47 m/s

Learn more about velocity:

brainly.com/question/5248528

#LearnwithBrainly

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Romashka [77]

A_x = 5.0

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Answer:

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