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3 years ago

Compared to its weight on Earth, a 5kg object on the moon will weigh

Physics

1 answer:

shutvik [7]3 years ago

5 0

Answer:

8.1 N/49 N=0.1653 which means 16.53% of the weight of the object on Earth.

Explanation:

On the Moon, where the gravitational constant is 1.62 $\frac{m}{s^2}$ , the weight of the 5 kg object will be: $weight_M=m*g_M = 5 kg * 1.62 \frac{m}{s^2} =8.1 N$

Where the answer is in Newtons (N) since all quantities are given in the SI system.

On Earth, on the other hand, the weight of the object is:

$weight_E=m*g_E= 5 kg* 9.8 \frac{m}{s^2} = 49N$

Therefore the object's weight on the Moon compared to that on Earth will be:

$8.1N/49N=0.1653$

That is, 16.53% of the weight the object has on Earth.

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At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.5 m/s2. At the same insta

Licemer1 [7]

Answer:

6.96 s

Explanation:

Given:

u = initial speed of the automobile = 0 m/s

a = constant acceleration of the automobile = $2.5\ m/s^2$
v = constant speed of the truck = 8.7 m/s

Assume:

t = time instant at which the automobile overtakes the truck.

At the moment the automobile and the truck both meat each other the distance travel by both vehicles must be the same.

$\therefore \textrm{Distance traveled by the automobile }=\textrm{Distance traveled by the truck}\\\Rightarrow ut+\dfrac{1}{2}at^2=vt\\\Rightarrow (0)t+\dfrac{1}{2}\times 2.5\times t^2=8.7t\\\Rightarrow 1.25t^2=8.7t\\\Rightarrow 1.25t^2-8.7t=0\\\Rightarrow t(1.25t-8.7)=0\\\Rightarrow t = 0\,\,\,or\,\,\, t = \dfrac{8.7}{1.25}\\\Rightarrow t = 0\,\,\,or\,\,\, t = 6.96\\$

Since t = 0 s is the initial condition. So, they both meet again at t = 6.96 s such that the automobile overtakes the truck.

6 0

3 years ago

How much support force does a table exert on a book that weighs 15 N when the book is placed on the table?

mestny [16]

Answer:

15 N

Explanation:

According to Newton's third law of motion, to every action, there is an equal and opposite reaction. This reaction is equal in magnitude to the force acting but in an opposite direction.

Now, if the book weighs 15 N, an opposite equal force will be: N = -15 N

But the magnitude of this will be the absolute value which is 15N.

8 0

2 years ago

An increase in the magnitude of velocity is??

bazaltina [42]

It means the speed of the object is increasing
and
there is a positive acceleration in the direction of the velocity
hence
there is a force acting on the object, in the direction of the velocity

8 0

3 years ago

A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc

marta [7]

This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so:

$v_{d} = \frac{J}{n\left | q \right |}$

The current I is any motion of charge from one region to another, so this is given by:

 $I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)$

The magnitude of the current density is:

$J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})$

Being:

$A=2mm^{2} = 2x10^{-6}m^{2}$

Finally, for the drift velocity magnitude vd, we find:

 $v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)$

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd

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2 years ago

You walk into a darkened room and turn on a flashlight. You see an image of the flashlight reflecting off a plane mirror in fron

user100 [1]

Answer:

4.2 is the answer

Explanation

The image formed in a plane mirror is an equal distance behind the mirror as the object in front of it.

Step 1: the equation to this problem would be: 8.4/2

Step 2: 8.4 ÷ 2 = 4.2

6 0

2 years ago