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3 years ago

A mass hanged on a spring scale. what is the force exerted by gravity on 700g ?

Physics

1 answer:

Ipatiy [6.2K]3 years ago

6 0

Answer:

6.86 N

Explanation:

Applying,

F = mg............... Equation 1

Where F = Force exerted by gravity on the mass, m = mass, g = acceleration due to gravity

Note: The Force exerted by gravity on the mass is thesame as the weight of the body.

From the question,

Given: m = 700 g = (700/1000) = 0.7 kg

Constant: g = 9.8 m/s²

Substitute these values into equation 1

F = 9.8(0.7)

F = 6.86 N

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A 20kg object acceleration by a force of 200N with coefficient of kineticfriction is 0.4 what is acceleration of the object?

Schach [20]

Answer:

Given - M = 20 kg

k = 0.4

F = 200 N

To find -  acceleration

Solution - 

F= kMA

200 = 0.4 * 20 * acceleration

200 = 8 * a

a = 8/200

a = 0.04 m s²

5 0

2 years ago

A solid yellow stripe is to be painted in the middle of a certain highway. If 1 gallon of paint covers an area of p square feet

marin [14]

Answer:

option A is correct

Explanation:

Given:

The length to be painted = m miles

The width to be painted = t inches

Area painted in 1 gallon = p square feet

Converting the every given dimension in feet, we have

length to be painted = m × 5280 feet

width to be painted = t/12 feet

area to be painted = (m × 5280 feet) × t/12 feet

now, applying the unitary method, we have

p square feet is painted ⇒ 1 gallon

1 square feet is painted ⇒ 1/p gallon

(m × 5280 feet) × t/12 feet square feet is painted ⇒ [(m × 5280 feet) × t/12 feet ] × 1/p gallon

thus, we get the gallons of paint required as 5820 mt/12p

hence option A is correct

3 0

3 years ago

A concave spherical mirror has a radius of curvature of magnitude 20.0 cm. (a) Find the location of the image for object distanc

Tems11 [23]

Answer:

Answered

Explanation:

The radius of curvature of the mirror R = 20 cm

then the focal length f = R/2 = 10 cm

(a) From mirror formula

1/f = 1/di + /1do

then the image distance

di = fd_o / d_o - f

= (10)(40) / 40-10

= 30.76 cm

since the image distance is positive so the image is real

ii) when the object distance d_0=20 cm

di = 10×20/ 20-10

= 20

Hence, the image must be real

iii)when the object distance d_0 = 10

di = 10×10 / 10-10 = ∞ (infinite)

the image will be formed at ∞

here also image will be real but diminished.

7 0

3 years ago

Read 2 more answers

Determine the critical crack length for a through crack contained within a thick plate of 7150-T651 aluminum alloy that is in un

Mila [183]

Explanation:

Formula to determine the critical crack is as follows.

$K_{IC} = \gamma \sigma_{f} \sqrt{\pi \times a}$

$\gamma$ = 1, $K_{IC}$ = 24.1

[/tex]\sigma_{y}[/tex] = 570

and, $\sigma_{f} = 570 \times \frac{3}{4}$

= 427.5

Hence, we will calculate the critical crack length as follows.

a = $\frac{1}{\pi} \times (\frac{K_{IC}}{\sigma_{f}})^{2}$

= $\frac{1}{3.14} \times (\frac{24.1}{427.5})^{2}$

= $10.13 \times 10^{-4}$

Therefore, largest size is as follows.

Largest size = 2a

= $2 \times 10.13 \times 10^{-4}$

= $20.26 \times 10^{-4}$

Thus, we can conclude that the critical crack length for a through crack contained within the given plate is $20.26 \times 10^{-4}$ .

4 0

3 years ago

A hot air balloon rising vertically is tracked by an observer located 3 miles from the lift-off point. At a certain moment, the

yuradex [85]

Answer:

$\frac{dy}{dt}=1.2\frac{mi}{min}$

Explanation:

We know that the tangent function relates the angle of the right triangle that forms the hot air balloon rising:

$tan\theta=\frac{y}{x}\\y=xtan\theta(1)$

Differentiating (1) with respect to time, we get:

$\frac{dy}{dt}=tan\theta\frac{dx}{dt}+xsec^{2}\theta\frac{d\theta}{dt}\\$

$\frac{dx}{dt}=0$ since x is a constant value. Replacing:

$\frac{dy}{dt}=3mi(sec^{2}\frac{\pi}{3})0.1\frac{rad}{min}\\\frac{dy}{dt}=1.2\frac{mi}{min}$

5 0

4 years ago