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3 years ago

If the textbook weight 19.6 newtons on venus, what is the strenght of gravity on that planet?

Mathematics

1 answer:

victus00 [196]3 years ago

6 0

Answer:

8.91m/s²

Step-by-step explanation:

If you were referring to the question :

If a 2.2 kg textbook weighs 19.6 newtons on Venus, what is the strength of gravity on that planet then read on.

Weight is the force of gravity on an object and it can be computed using the formula:

W = mg

where:

W = weight (N)

m = mass (kg)

g = acceleration due to gravity

So we know the weight and the mass of the object, all we do is plug it into our formula and solve for what we do not know.

$W = mg\\\\19.6N = (2.2kg)(g)\\\\ \dfrac{16.6N}{2.2kg} = g\\\\8.91m/s^{2} = g$

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Bags of pretzels are sampled to ensure proper weight. The overall average for the samples is 9 ounces. Each sample contains 25 b

77julia77 [94]

Answer:

The value is $UCL = 10.8$

Step-by-step explanation:

From the question we are told that

The sample mean is $\= x = 9 \ ounce$

The sample size is n = 25

The standard deviation is $\sigma = 3 \ ounce$

Given that the sample size is not large enough i.e n< 30 we will make use of the student t distribution table

From the question we are told the confidence level is 99.7% , hence the level of significance is

$\alpha = (100 -99.7 ) \%$

=> $\alpha = 0.003$

Generally the degree of freedom is $df = n- 1$

=> $df = 25 - 1$

=> $df = 24$

Generally from the student t distribution table the critical value of $\frac{\alpha }{2}$ at a degree of freedom of $df = 24$ is

$t_{\frac{\alpha }{2} , 24 } = 3.0$

Generally the margin of error is mathematically represented as

$E = t_{\frac{\alpha }{2} , 24} * \frac{\sigma }{\sqrt{n} }$

=> $E = 3.0 * \frac{3 }{\sqrt{25} }$

=> $E =1.8$

Gnerally the upper control chart limit for 99.7% confidence is mathematically represented as

$UCL = \= x + E$

=> $UCL = 9 + 1.8$

=> $UCL = 10.8$

4 0

2 years ago

3/10 minus what equals 2/5?

Jlenok [28]

7/10-3/10=2/5
3/10+1/10=2/5

3 0

3 years ago

Read 2 more answers

Compare 6. 108 to 3 . 106<br> How much bigger is it from each other

vlada-n [284]

3.002
To solve this problem, all you need to do is subtract 6.108-3.106

5 0

3 years ago

Read 2 more answers

Which expression is equivalent to

Scilla [17]

The equivalent expression of $\frac{(5ab)^3}{30ab^{-7}}$ is $\frac{25a^{2}b^{10}}{6}$

<h3>How to determine the equivalent expression?</h3>

The expression is given as:

$\frac{(5ab)^3}{30ab^{-7}}$

Expand the numerator

$\frac{(5ab)^3}{30ab^{-7}} =\frac{125a^3b^3}{30ab^{-7}}$

Divide 125 and 30 by 5

$\frac{(5ab)^3}{30ab^{-7}} =\frac{25a^3b^3}{6ab^{-7}}$

Apply the law of indices

$\frac{(5ab)^3}{30ab^{-7}} =\frac{25a^{3-1}b^{3+7}}{6}$

Evaluate the exponent

$\frac{(5ab)^3}{30ab^{-7}} =\frac{25a^{2}b^{10}}{6}$

Hence, the equivalent expression of $\frac{(5ab)^3}{30ab^{-7}}$ is $\frac{25a^{2}b^{10}}{6}$

Read more about equivalent expression at:

brainly.com/question/2972832

#SPJ1

6 0

2 years ago

what are the coordinates of point T’, the image of point T(-2,5) after a rotation of 180 about the origin

RideAnS [48]

Answer:

T'(2, - 5 )

Step-by-step explanation:

Under a rotation about the origin of 180°

a point (x, y ) → (- x, - y ), thus

T(- 2, 5 ) → T'(2, - 5 )

5 0

3 years ago