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2 years ago

6

Different masses are hung on a spring scale calibrated in newtons. The force exerted by gravity on 1.0 kg is shown in the image

below in newtons.
An image of a spring scale with a 1 kilogram weight attached. The scale reads 9.8 N.

What would be the force exerted by gravity on 6.5 kg?

6.5 N
9.8 N
64 N
98 N

2 answers:

Lunna [17]2 years ago

6 0

6.5 x 9.8
=63.7
so the answer is 64N

Arada [10]2 years ago

3 0

Answer:

F = 64 N

Explanation:

When 1 kg of weight attached with the spring, the scale reads 9.8 N. Weight of an object is the product of mass and acceleration due to gravity. Let F is the force exerted by gravity on 6.5 kg. It can be calculated as :

$F=mg$

$F=6.5\ kg\times 9.8\ m/s^2$

F = 63.7 N

or

F = 64 N

So, the force exerted by gravity on 6.5 kg is 64 N.

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When you are on a huge water slide what forces are there? when will you experience a net force?

Pani-rosa [81]

When you are on a huge water slide, the force present as you slide is the gravitational force. It is because the gravity enables you to slide down the water slide. The net force is the overall forces of the object, so as you slide the water slide, you may experience the net force once you slide down with the gravity and water sliding you down.

3 0

3 years ago

If a force of 19 N is applied to a crate to push it 23 m, how much work is done on the crate by the force?

swat32

Answer:

437 Joules

Explanation:

Use the formula for work directly

(work) = (force) x (displacement)

to get

(work) = (19 N) x (23 m) = 437 Joules

4 0

3 years ago

A wire of length 5mm and Diameter 2m extends by 0.25 when a force of 50N was use. calculate the

bazaltina [42]

Answer and Explanation:

Data provided in the question

Force = 50N

Length = 5mm

diameter = 2.0m = $2\times 10^{-3}$

Extended by = 0.25mm = $0.25\times 10^{-3}$

Based on the above information, the calculation is as follows

a. The Stress of the wire is

$= \frac{force\ applied}{area\ of \ circle}$

here area of circle = perpendicular to the are i.e cross-sectional i.e

= $\frac{\pi d^{2}}{4}$

= $\frac{\pi(2\times 10^{-3})^2}{4}$

Now place these above values to the above formula

$= \frac{4\times 50}{\pi\times 4 \times 10^{-6}} \\\\ = \frac{50}{\pi}$

= 15.92 MPa

As 1Pa = 1 by N m^2

So,

MPa = 10^6 N m^2

b. Now the strain of the wire is

$= \frac{Change\ in\ length}{initial\ length} \\\\ = \frac{0.25\times 10^{-3}}{5}$

= $5 \times 10^{-5}$

3 0

3 years ago

A golf ball comes to rest 1.90 seconds after hitting a net. The force of the net slowed the ball down at a rate of 56.3 m/s^2. W

Nadusha1986 [10]

Answer:

The velocity of the ball when it first hit the net is Vi= 106.97 m/s

Explanation:

Vf= 0

Vi= ?

a= 56.3 m/s²

t= 1.9 s

Vf= Vi - a*t

Vi= a*t

Vi= 106.97 m/s

8 0

3 years ago

Two ice skaters, Daniel (mass 70.0 kg) and Rebecca (mass 45.0 kg), are practicing. Daniel stops to tie his shoelace and, while a

wel

Answer:

a) $v=7.32m/s$

b) $\alpha =-35$ º

c) $ΔK=-1094.62J$

Explanation:

From the exercise we know that the collision between Daniel and Rebecca is elastic which means they do not stick together

So, If we analyze the collision we got

$p_{1x}=p_{2x}$

To simplify the problem, lets name D for Daniel and R for Rebecca

a) $p_{D1x}+p_{R1x}=p_{D2x}+p_{R2x}$

Since Daniel's initial velocity is 0

$m_{R}v_{Rx}=m_{D}v_{D2x}+m_{R}v_{R2x}$

$v_{D2x}=\frac{m_{R}*v_{R1x}-m_{R}*v_{R2x}}{m_{D}}$

$v_{D2x}=\frac{(45kg)(14m/s)-(45kg)(8cos(55.1)m/s)}{(70kg)}=6m/s$

Now, lets analyze the movement in the vertical direction

$p_{1y}=p_{2y}$

Since $p_{1y}=0$

$0=m_{D}v_{D2y}+m_{R}v_{R2y}$

$v_{D2y}=-\frac{m_{R}v_{2Ry}}{m_{D}}=-\frac{(45kg)(8sin(55.1)m/s)}{(70kg)}=-4.21m/s$

Now, we can find the magnitude of Daniel's velocity after de collision

$v_{D}=\sqrt{(6m/s)^2+(-4.21m/s)^2}=7.32m/s$

b) To know whats the direction of Daniel's velocity we need to solve the arctan of the angle

$\alpha =tan^{-1}(\frac{v_{y}}{v_{x}})=tan^{-1}(\frac{-4.21}{6})=-35$ º

c) The change in the total kinetic energy is:

ΔK= $K_{2}-K_{1}$

ΔK= $\frac{1}{2}[(45kg)(8m/s)^2+(70kg)(7.32m/s)^2-(45kg)(14m/s)^2]=-1094.62J$

That means that the kinetic energy decreases

5 0

2 years ago