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3 years ago

A 20kg suitcase on a flat surface is pushed by a force of 100N downwards at an

Physics

1 answer:

uranmaximum [27]3 years ago

8 0

Answer:

282.6 N

Explanation:

Draw a free body diagram of the suitcase. There are three forces:

Weight force mg pulling down,

Normal force N pushing up,

and 100 N down at angle 60° with the horizontal.

Sum of forces in the y direction:

∑F = ma

N − mg − 100 sin 60° = 0

N = mg + 100 sin 60°

N = (20 kg) (9.8 m/s²) + 100 sin 60°

N = 282.6 N

Round as needed.

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A golf club rotates 215 degrees and has a length (radius) equal to 29 inches. The time it took to swing the club was 0.8 seconds

vichka [17]

Answer:

The average linear velocity (inches/second) of the golf club is 136.01 inches/second

Explanation:

Given;

length of the club, L = 29 inches

rotation angle, θ = 215⁰

time of motion, t = 0.8 s

The angular speed of the club is calculated as follows;

$\omega = (\frac{\theta}{360} \times 2\pi, \ rad) \times \frac{1}{t} \\\\\omega = (\frac{215}{360} \times 2\pi, \ rad) \times \frac{1}{0.8 \ s} \\\\\omega = 4.69 \ rad/s$

The average linear velocity (inches/second) of the golf club is calculated as;

v = ωr

v = 4.69 rad/s x 29 inches

v = 136.01 inches/second

Therefore, the average linear velocity (inches/second) of the golf club is 136.01 inches/second

8 0

3 years ago

A cruise ship sails due south at 2.00 m/s while a coast guard patrol boat heads 19.0° north of east at 5.60 m/s. What are the x-

Lilit [14]

Answer:

The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

Explanation:

Given that,

Velocity of ship = 2.00 m/s due south

Velocity of boat = 5.60 m/s due north

Angle = 19.0°

We need to calculate the component

The velocity of the ship in term x and y coordinate

$v_{s_{x}}=0$

$v_{s_{y}}=2.0\ m/s$

The velocity of the boat in term x and y coordinate

For x component,

$v_{b_{x}}=v_{b}\cos\theta$

Put the value into the formula

$v_{b_{x}}=5.60\cos19$

$v_{b_{x}}=5.29\ m/s$

For y component,

$v_{b_{y}}=v_{b}\sin\theta$

Put the value into the formula

$v_{b_{y}}=5.60\sin19$

$v_{b_{y}}=1.82\ m/s$

We need to calculate the x-component and y-component of the velocity of the cruise ship relative to the patrol boat

For x component,

$v_{sb_{x}}=v_{s_{x}}-v_{b_{x}}$

Put the value into the formula

$v_{sb_{x}=0-5.29$

$v_{sb}_{x}=-5.29\ m/s$

For y component,

$v_{sb_{y}}=v_{s_{y}}-v_{b_{y}}$

Put the value into the formula

$v_{sb_{x}=2.-1.82$

$v_{sb}_{x}=0.18\ m/s$

Hence, The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

7 0

3 years ago

Where was land-dwelling found?

Sav [38]

Remote Scottish island

4 0

4 years ago

Read 2 more answers

An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0

Murrr4er [49]

Complete question:

An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0 V battery. Find the value of its capacitance.

Answer:

The value of its capacitance is 1.027 x 10⁻¹² F

Explanation:

Given;

area of the plate, A = 2.9 cm² = 2.9 x 10⁻⁴ m²

separation distance of the plates, d = 2.5 mm = 2.5 x 10⁻³ m

voltage of the battery, V = 18 V

The value of its capacitance is calculated as;

$C = \frac{k\epsilon_0A}{d} \\\\C = \frac{(1)(8.85\times 10^{-12})(2.9 \times 10^{-4})}{2.5 \times 10^{-3}} \\\\C = 1.027 \times 10^{-12} \ F$

Therefore, the value of its capacitance is 1.027 x 10⁻¹² F

8 0

3 years ago

"In a Young’s double-slit experiment, the separation between slits is d and the screen is a distance D from the slits. D is much

san4es73 [151]

Answer:

The number of bright fringes per unit width on the screen is, $x=\dfrac{\lambda D}{d}$

Explanation:

If d is the separation between slits, D is the distance between the slit and the screen and $\lambda$ is the wavelength of the light. Let x is the number of bright fringes per unit width on the screen is given by :

$x=\dfrac{n\lambda D}{d}$

$\lambda$ is the wavelength

n is the order

If n = 1,

$x=\dfrac{\lambda D}{d}$

So, the the number of bright fringes per unit width on the screen is $\dfrac{\lambda D}{d}$ . Hence, the correct option is (B).

6 0

3 years ago