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3 years ago

Describe the motion represented by a horizontal line on a distance-time graph.

Physics

1 answer:

denpristay [2]3 years ago

3 0

Answer:

hi, this is the answer

Explanation:

A horizontal line on a distance-time graph shows no change in distance, therefore there is no motion.

The object is stationary. ...

Constant speed is motion that occurs with the same ratio of distance to time throughout the entire length of the motion.

pls mark this as the brainliest...

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Eq-48 when operating your vessel with a vhf radio, what channel must you monitor?

77julia77 [94]

VHF Channel 16 must always be monitored. It is the calling and distress channel. This is where the operator might hear a MAYDAY call. This is also where the operator can issue a MAYDAY. Improper misuse of the channel can result to penalties.

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3 years ago

A stagehand starts sliding a large piece of stage scenery originally at rest by pulling it horizontally with a force of 176 N. W

Diano4ka-milaya [45]

Answer:

Option (a)

Explanation :

A stage hand starts sliding a large piece of stage scenery originally at rest by pulling it horizontally with a force of 176 N.

Hence Force applied $\text { Fapplied }=176 \mathrm{N}$

Force on piece of scenery $\mathrm{F}_{\mathrm{g}}=490 \mathrm{N}$

$\Sigma \mathrm{F} \mathrm{y}=\mathrm{F} \mathrm{n}-\mathrm{Fg}=0$

$\mathrm{Fn}=\mathrm{Fg}$

$\mathrm{FS}_{\mathrm{max}}=\mathrm{F}_{\mathrm{applied}}$

µk = $\frac{\mathrm{Fs} \max }{\mathrm{Fn}}$

= $\frac{\text { Fapplied }}{\mathrm{Fg}}$

= $\frac{176}{490}$ =0.36

coefficient of static friction is 0.36

6 0

4 years ago

A 1250-kg compact car is moving with velocity v1 =36.2i^+12.7j^m/s. It skids on a frictionless icy patch and collides with a 448

MA_775_DIABLO [31]

Momentum is conserved, so the sum of the separate momenta of the car and wagon is equal to the momentum of the combined system:

(1250 kg) ((36.2 i + 12.7 j ) m/s) + (448 kg) ((13.8 i + 10.2 j ) m/s) = ((1250 + 448) kg) v

where v is the velocity of the system. Solve for v :

v = ((1250 kg) ((36.2 i + 12.7 j ) m/s) + (448 kg) ((13.8 i + 10.2 j ) m/s)) / (1698 kg)

v ≈ (30.3 i + 12.0 j ) m/s

7 0

3 years ago

What happens at the Bernoulli effect

Paul [167]

Answer:

Within a horizontal flow fluid, points of higher fluid speed will have less pressure than points of slower fluid speed.

Explanation:

5 0

3 years ago

A 70-cm-diameter wheel accelerates uniformly from 160rpm to 280rpm in 4.0s. Determine (a) its angular acceleration, and (b) the

poizon [28]

Explanation:

Given that,

Diameter of the wheel, d = 70 cm = 0.7 m

Initial angular speed, $\omega_i=160\ rpm=16.75\ rad/s$

Final angular speed, $\omega_f=280\ rpm=29.32\ rad/s$

Time, t = 4 s

(a) Angular acceleration,

$\alpha =\dfrac{\omega_f-\omega_i}{t}\\\\\alpha =\dfrac{29.32-16.75}{4}\\\\\alpha =3.14\ rad/s^2$

(b) Tangential acceleration is :

$a=r\alpha \\\\a=0.35\times 3.14\\\\a=1.085\ m/s^2$

Angular speed of the wheel after 2 seconds is :

$\omega_f=\omega_i+\alpha t\\\\\omega_f=16.75+3.14\times 2\\\\\omega_f=23.03\ rad/s$

Radial acceleration will be :

$a=\omega_f^2r\\\\a=(23.03)^2\times 0.35\\\\a=185.6\ m/s^2$

Hence, this is the required solution.

4 0

3 years ago