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3 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]3 years ago

6 0

6.5 x 9.8
=63.7
so the answer is 64N

Arada [10]3 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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Why is speed a scaler quantity

In-s [12.5K]

Answer:

Speed is a scalar quantity it is the rate of change in the distance travelled by an object, while velocity is a vector quantity it is the speed of an object in a particular direction.

4 0

4 years ago

A horizontal force of 40N is needed to pull a 60kg box across the horizontal floor at which coefficient of friction between floo

Mazyrski [523]

Answer:

Coefficient of friction is $0.068$ .

Work done is $320~J$ .

Explanation:

Given:

Mass of the box ( $m$ ): $60$ kg

Force needed ( $F$ ): $40$ N

The formula to calculate the coefficient of friction between the floor and the box is given by

$F=\mu mg...................(1)$

Here, $\mu$ is the coefficient of friction and $g$ is the acceleration due to gravity.

Substitute $40$ N for $F$ , $60$ kg for $m$ and $9.80$ m/s² for $g$ into equation (1) and solve to calculate the value of the coefficient of friction.

$40 N=\mu\times60 kg\times9.80 m/s^{2} \\~~~~~\mu=\frac{40 N}{60 kg\times9.80 m/s^{2}}\\~~~~~~~=0.068$

The formula to calculate the work done in overcoming the friction is given by

$W=Fd..........................(2)$

Here, $W$ is the work done and $d$ is the distance travelled.

Substitute $40$ N for $F$ and $8 m$ for $d$ into equation (2) to calculate the work done.

$W=40~N\times8~m\\~~~~= 320~J$

8 0

3 years ago

A student decides to give his bicycle a tune up. He flips it upside down (so there's no friction with the ground) and applies a

aleksklad [387]

Answer:

$V=9.2565m/s$

Explanation:

From the question we are told that:

Force $F = 34 N$

Time $t = 0.6 s$

Length of pedal $l_p=16.5cm \approx0.165m$

Radius of wheel $r = 33 cm = 0.33 m$

Moment of inertia, $I = 1200 kgcm2 = 0.12 kg.m2$

Generally the equation for Torque on pedal $\mu$ is mathematically given by

$\mu=F*L\\\mu=34*0.165$

$\mu=5.61N.m$

Generally the equation for angular acceleration $\alpha$ is mathematically given by

$\alpha=\frac{\mu}{l}$

$\alpha=\frac{5.61}{0.12}$

$\alpha=46.75$

Therefore Angular speed is \omega

$\omega=\alpha*t$

$\omega=(46.75)*(0.6)$

$\omega=28.05rad/s$

Generally the equation for Tangential velocity V is mathematically given by

$V=r\omega$

$V=(0.33)(28.05)$

$V=9.2565m/s$

5 0

3 years ago

Describe how light is used in modern forms of communication.

Kobotan [32]

Answer:Visible light communication (VLC) is a wireless method that uses light emitted by LEDs to deliver networked, mobile, high-speed communication similar to Wi-Fi, leading to the term Li-Fi. It can be used as standalone solution or in a supplementary role to radio-frequency (RF) or cellular network communication

Explanation:

8 0

3 years ago

Suppose a straight 1.00-mm-diameter copper (density = 8.9 x103 kg/m3) wire could just “float” horizontally in air because of the

deff fn [24]

We are given
diameter, d = 1.00 mm
density, ρ = 8.9 x10^3 kg/m3
magnetic field, B = 5.0 x 10^-5 T

We are asked to determine the current that would be carried by the wire
We use the formula
B = I R / A
where I is the current
R is the resistance and R = πρd / A
So,
B = I (πρd /A) / A
B = I πρd / A²

and
A is the area, A = πd²/4
So,
B = I πρd / (πd²/4)²
B = I πρd / (π² d⁴ / 16)
B = 16 I ρ / πd²

Substituting the given and solving for I
5.0 x 10^-5 = 16 I (8.9 x10^3 kg/m3) / π (1 x 10^-3)²
I = 1.1 x 10^-15 A

4 0

4 years ago