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3 years ago

Describe friction force (Ffr) in your own words

1 answer:

3 0

Answer: Friction is like when u take two stick's and rub it together to make fire when you use friction it can produce heat. Applied force is like and applied to an object or person if a person is pushing a desk across the room then there is applied force.

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A 1056-hertz tuning fork is sounded at the same time a piano note is struck. You hear three beats per second. What is the freque

REY [17]

Answer:

$f_2 = 1053 Hz$

$f_2 = 1059 Hz$

Explanation:

As we know that the beat frequency is given as

$|f_1 - f_2| = f_{beat}$

here we know that

$f_1 = 1056 Hz$

$f_{beat} = 3 Hz$

now we have

$|1056 - f_2| = 3$

now we have

$1056 - f_2 = \pm 3$

so we have

$f_2 = 1053 Hz$

$f_2 = 1059 Hz$

5 0

4 years ago

Select the object that has the most gravitational force. a sand grain a marble a bowling ball a tennis ball

Anna35 [415]

<span>So we want to know which object has a gravitational force of the greatest magnitude. Since gravitational force depends on the mass of the object, the object with the greatest mass will have the greatest magnitude of the gravitational force. In this case that is the bowling ball. </span>

7 0

3 years ago

Read 2 more answers

Using a refracting telescope, you observe the planet Mars when it is 2.11 × 10 11 m 2.11×1011 m from Earth. The diameter of the

Norma-Jean [14]

Answer:

y = 150 Km

Explanation:

given,

Distance of the planet Mars from earth D = 2.11 × 10¹¹ m

The diameter of the telescope's objective lens d = 0.949 m

The wavelength of light λ = 553 nm

= 553 x 10⁻⁹ m

resolution of the telescope = ?

$y = \dfrac{1.22\lambda\ D}{d}$

$y = \dfrac{1.22\times 553\times 10^{-9}\times 2.11 \times 10^{11}}{0.949}$

y = 150003 m

y = 150 Km

the minimum feature size is equal to y = 150 Km

8 0

3 years ago

A cylindrical rod of mass M. length L and radius R has two cords wound around it whose ends are a

Rudik [331]

Answer:

T = mg/6

Explanation:

Draw a free body diagram (see attached). There are two tension forces acting upward at the edge of the cylinder, and weight at the center acting downwards.

The center rotates about the point where the cords touch the edge. Sum the torques about that point:

∑τ = Iα

mgr = (1/2 mr² + mr²) α

mgr = 3/2 mr² α

g = 3/2 r α

α = 2g / (3r)

(Notice that you have to use parallel axis theorem to find the moment of inertia of the cylinder about the point on its edge rather than its center.)

Now, sum of the forces in the y direction:

∑F = ma

2T − mg = m (-a)

2T − mg = -ma

Since a = αr:

2T − mg = -mαr

Substituting expression for α:

2T − mg = -m (2g / (3r)) r

2T − mg = -2/3 mg

2T = 1/3 mg

T = 1/6 mg

The tension in each cord is mg/6.

7 0

3 years ago

A 50-turn circular coil with radius 4 cm is placed in magnetic field of 6000 G. An amount of 800 mA current passes through the c

satela [25.4K]

Explanation:

n=50,r=0.02mn=50,r=0.02m,

I=5AandB=0.20TI=5AandB=0.20T

τismaxiμmwhensinθ=90∘τismaxiμmwhensinθ=90∘

τmax=niabsin90∘=mbτmax=niabsin90∘=mb

=50×5×3.14×4×10−4×2×10−1=50×5×3.14×4×10-4×2×10-1

=6.28×10−2Nm=6.28×10-2Nm

Given τ=12×τmaxτ=12×τmax

⇒sinsinθ=12⇒sinsinθ=12 or sinθ=30∘sinθ=30∘

=∠betweenareavar→randmag≠ticfield=∠betweenareavar→randmag≠ticfield.

So angle between magnetic field and the plane of the coil

=90∘−30∘=60∘=90∘-30∘=60∘.

<h3>HOPE IT HELPS </h3>

<h2>mark me in brainliest answers please please please </h2>

4 0

3 years ago