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3 years ago

8

A block weighing 15N rest on a flat surface and a horizontal force of 3N is exerted on it. Determine the frictional force on the

block

1 answer:

Ipatiy [6.2K]3 years ago

7 0

The answer you are looking for is 3.0N

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Bismuth-210 has a half-life of 5 days. If you

Anton [14]

Answer:

A = 62.5 g

using half life formulas, plug in missing variables.

6 0

4 years ago

A bullet is fired with a muzzle velocity of 1178 ft/sec from a gun aimed at an angle of 26° above the horizontal. Find the horiz

dalvyx [7]

Answer:

1058.78 ft/sec

Explanation:

Horizontal Component of Velocity; This is the velocity of a body that act on the horizontal axis. I.e Velocity along x-axis

The horizontal velocity of a body can be calculated as shown below.\

Vh = Vcos∅.......................... Equation 1

Where Vh = horizontal component of the velocity, V = The velocity acting between the horizontal and the vertical axis, ∅ = Angle the velocity make with the horizontal.

Given: V = 1178 ft/sec, ∅ = 26°

Substitute into equation 1

Vh = 1178cos26

Vh = 1178(0.8988)

Vh = 1058.78 ft/sec

Hence the horizontal component of the velocity = 1058.78 ft/sec

8 0

3 years ago

marishachu [46]

Answer:

The two moments must be the same:

p1=p2

m1v1=m2v2

v2=(m1v1)/m2

v2=(90 kg x 0.9 m/s)/110kg=0.7 m/s

3 0

3 years ago

A yo‑yo with a mass of 0.0800 kg and a rolling radius of =2.70 cm rolls down a string with a linear acceleration of 5.70 m/s2.

N76 [4]

Explanation:

Given that,

Mass, m = 0.08 kg

Radius of the path, r = 2.7 cm = 0.027 m

The linear acceleration of a yo-yo, a = 5.7 m/s²

We need to find the tension magnitude in the string and the angular acceleration magnitude of the yo‑yo.

(a) Tension :

The net force acting on the string is :

ma=mg-T

T=m(g-a)

Putting all the values,

T = 0.08(9.8-5.7)

= 0.328 N

(b) Angular acceleration,

The relation between the angular and linear acceleration is given by :

$\alpha =\dfrac{a}{r}\\\\\alpha =\dfrac{5.7}{0.027}\\\\=211.12\ m/s^2$

(c) Moment of inertia :

The net torque acting on it is, $\tau=I\alpha$ , I is the moment of inertia

Also, $\tau=Fr$

So,

$I\alpha =Fr\\\\I=\dfrac{Fr}{\alpha }\\\\I=\dfrac{0.328\times 0.027}{211.12}\\\\=4.19\times 10^{-5}\ kg-m^2$

Hence, this is the required solution.

3 0

3 years ago

Steve is recording the temperature of an object on Celsius and Fahrenheit scales. Relationship between the two scales is given b

Vlad [161]

Answer:

False

Explanation:

7 0

3 years ago