All categories

3 years ago

An object has a mass of 20 kg. Determine its weight, in N, at a location where the acceleration of gravity is 9.78 m/s2.

Physics

1 answer:

pychu [463]3 years ago

3 0

Answer:

Weight of the object will be 195.6 N

Explanation:

We have given mass of the object m = 20 kg

And acceleration due to gravity at any location is given $g=9.78m/sec^2$

We have to find the weight of the object in N at the location where value of acceleration due to gravity is $g=9.78m/sec^2$

Weight of the object is given by W = mg , here W is weight , m is mass and g is acceleration due to gravity

So weight W = 20×9.78 = 195.6 N

So weight of the object will be 195.6 N

You might be interested in

WILL MARK AS BRAINLIEST IF THE ANSWER IS CORRECT!!!

Bogdan [553]

The answer is D. Current.

7 0

3 years ago

Read 2 more answers

A metal wire of resistance 2.30 10-8 is cut into two equal pieces that are then connected side by side to form a new wire the le

Elena-2011 [213]

Equal parallel resistors are governed by the formula

R = r^2/(2r) = r/2

r = 2.30 * 10^-8 ohms which is very tiny.

R = 2.30 *10^-8 / 2

R = 1.15 * 10^-8 ohms. <<<< Answer

7 0

3 years ago

A city planner is working on the redesign of a hilly portion of a city. An important consideration is how steep the roads can be

Artyom0805 [142]

Explanation:

It is known that relation between force and acceleration is as follows.

F = $m \times a$

I is given that, mass is 1090 kg and acceleration is 21 m/s. Therefore, we will calculate force as follows.

F = $m \times a$

= $1090 \times (\frac{21}{16})$

= 1430.625 N

Also, it is known that

$sin(\theta) = \frac{\text{Force car can exert}}{\text{Force gravity pulls car}}$

$sin(\theta) = \frac{1430.625 N}{(1090 \times 9.8) N}$

$\theta$ = 7.70 degrees

Thus, we can conclude that the maximum steepness for the car to still be able to accelerate is 7.70 degrees.

8 0

3 years ago

What is the relationship between the intervals on the Celsius scale and the Kelvin scale

MAXImum [283]

An interval of 1 degree Celsius is equal to 1 Kelvin

Explanation:

There are different scales used for measuring temperatures. The most common ones are:

- Celsius scale: it is set by the following convention: the zero in this scale is placed at the freezing temperature of the water, while the 100 degrees is placed in corresponding to the boiling point of water. The the scale is divided into 100 equal intervals between these two points. Temperatures in this scale are indicated with the Celsius degree ( $^\circ{}C$ .

- Kelvin scale: it is set such that the absolute zero (the temperature at which particles in matter stop to have any motion) is placed at zero in this scale. A difference of 1 degree in this scale is equal to 1 degree Celsius. Temperatures in this scale are indicated with the Kelvin (K).

The conversion between the two temperatures is:

$T(^{\circ}) = T(K) - 273.15$

So for example, the absolute zero ( $0 K$ ) corresponds to $-273.15^{\circ}C$ in the Celsius scale. However, intervals between two temperatures are measured in the same way in the two scales: therefore, an interval of 1 degree in the Celsius scale is equal to an interval of 1 Kelvin in the Kelvin scale.