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Pani-rosa [81]
3 years ago
6

Anuja is holding a stuffed dog, with a mass of 0.30 kg, when Derek decides that he wants it and tries to pull it away from Abuja

. If Derek pulls horizontally on the dog with a force of 10.0 N and Anudja pulls with a horizontal force of 11.0 N, what is the horizontal acceleration of the dog?
Physics
1 answer:
igor_vitrenko [27]3 years ago
8 0

Answer:

a = 3.33 m/s²

Explanation:

The horizontal acceleration of the dog can be found by using Newton's Second Law of Motion as follows:

F = ma

where,

F = Unbalanced force applied on the dog = 11 N - 10 N = 1 N

m = mass of the dog = 0.3 kg

a = horizontal acceleration of dog = ?

Therefore,

1 N = 0.3 kg(a)

a = 1 N/0.3 kg

<u>a = 3.33 m/s²</u>

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I know this may not be the answer youre looking for, but hopefully this can help somehow!

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3 years ago
ECONOMICS GRADE 10 CASE STUDY TOPIC: South African growth and development: Mining and industry Manufacturing and services.​
bearhunter [10]

The correct answe is South Africa - Economic Growth and Development:

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One of the largest economies on the African continent is that of South Africa. However, despite a period of rapid development from 2003 to 2007, its real GDP average yearly growth rate between 2001 and 2010 has been very weak and unquestionably significantly below the African average. A number of African nations have had substantially faster growth rates, which has improved a number of development-related indices.

By the standards of the recent growth records of several Euro Zone nations, South Africa's development is hardly sluggish! One crucial aspect of the economy is that South Africa has achieved relatively modest progress in meeting a number of important development targets and some of the Millennium Development Goals, but her economy does not appear to have achieved the "take-off" required to kick start significant development progress, especially against the backdrop of her deep social problems.

To learn more about South Africa's development refer the link:

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6 0
2 years ago
A toroid having a square cross section, 5.49 cm on a side, and an inner radius of 18.1 cm has 450 turns and carries a current of
joja [24]

Answer:

(a) 4.27 x 10^-4 Telsa

(b) 3.28 x 10^-4 Telsa

Explanation:

side of square, a = 5.49 cm

inner radius, r = 18.1 cm = 0.181 m

number of turns,N = 450

current, i = 0.859 A

(a)

The magnetic field due to a solenoid due to inner radius is

B = \frac{\mu_{0}Ni}{2\pi r_{inner}}

B = \frac{4\pi\times 10^{-7}\times 450\times 0.859}{2\pi \times 0.181}

B = 4.27 x 10^-4 Telsa

(b)

The outer radius is R = 18.1 + 5.49 = 23.59 cm = 0.236 m

The magnetic field due to the outer radius is

B = \frac{\mu_{0}Ni}{2\pi r_{outer}}

B = \frac{4\pi\times 10^{-7}\times 450\times 0.859}{2\pi \times 0.236}

B = 3.28 x 10^-4 Tesla

7 0
4 years ago
The closest distance a book can be read from a pair of reading eyeglasses (Power = 1.55 dp) is 26.0 cm. What is the near distanc
Mnenie [13.5K]

Answer:

The image distance is 20.0 cm.

Explanation:

Given that,

Power = 1.55 dp

Distance between book to eye = 26.0+3.00=29.0 cm

We need to calculate the focal length

Using formula of focal length

f = \dfrac{1}{P}

Put the value into the formula

f=\dfrac{1}{1.55}

f=0.645\ m

f=64.5\ cm

We need to calculate the image distance

Using lens formula

\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}

\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{-u}

Put the value into the formula

\dfrac{1}{v}=\dfrac{1}{64.5}-\dfrac{1}{-29}

\dfrac{1}{v}=\dfrac{187}{3741}

v=20.0\ cm

Hence, The image distance is 20.0 cm.

5 0
3 years ago
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