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3 years ago

Which if the following doesn't cause an object to accelrate it's q 4

Physics

2 answers:

GarryVolchara [31]3 years ago

7 0

Change in direction, slowing down, and speeding up are all CHANGES of velocity, and THAT's called acceleration.

Constant velocity isn't a change of velocity, so it isn't called acceleration.

Slav-nsk [51]3 years ago

6 0

B. Constant velocity

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A linear network has a current input 7.5 cos(10t + 30°) A and a voltage output 120 cos(10t + 75°) V. Determine the associated im

Leona [35]

Answer:

16∠45° Ω

Explanation:

Applying,

Z = V/I................... Equation 1

Where Z = Impedance, V = Voltage output, I = current input.

Given: V = 120cos(10t+75°), = 120∠75°, I = 7.5cos(10t+30) = 7.5∠30°

Substitute these values into equation 1

Z = 120cos(10t+75°)/7.5cos(10t+30)

Z = 120∠75°/ 7.5∠30°

Z = 16∠(75°-30)

Z = 16∠45° Ω

Hence the impedance of the linear network is 16∠45° Ω

8 0

3 years ago

A 1350 kg car travels at 12 m/s. What is it's kinetic energy

Lana71 [14]

Hello!

A 1350 kg car travels at 12 m/s. What is it's kinetic energy ?

We have the following data:

m (mass) = 1350 kg

v (velocity) = 12 m/s

KE (kinetic Energy) = ? (in Joule)

Formula to calculate kinetic energy:

$\boxed{KE = \dfrac{1}{2} *m*v^2}$

Solving:

$KE = \dfrac{1}{2} *m*v^2$

$KE = \dfrac{1}{2} *1350*12^2$

$KE = \dfrac{1}{2} *1350*144$

$KE = \dfrac{194400}{2}$

$\boxed{\boxed{KE = 97200\:Joule}}\:or\:\boxed{\boxed{KE = 97.2\:kJ}}\Longleftarrow(kinetic\:energy)\:\:\:\:\:\:\bf\blue{\checkmark}$

_______________________

$\bf\green{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}$

5 0

3 years ago

Help!!

Ulleksa [173]

Answer:

5.5 KG is the mass

53.9 is the Weight

5 0

3 years ago

A student attempts to slide a block down a smooth, dry ramp as shown, but the block does not move. The most likely explanation i

Rashid [163]

Answer:

Explanation:

The normal force would be that the ramp would be dry

6 0

3 years ago

It you balance a weight of 20 N at the 15 cm MARK and want to balance a weight of on the other side of the meter stick at 70 cm

agasfer [191]

Answer:

The weight that will balance the meter stick at 70 cm mark is 35N

Explanation:

Given;

first weight at 15 cm mark, W₁ = 20 N

second weight at 70 cm mark, W₂ = ?

A sketch of the question in diagram form;

The center of gravity of the meter stick is 50 cm

15cm 50cm 70cm

0---------------------------------------Δ------------------------------------------100cm

↓ ↓

20N W₂

<------------------>|<--------------->

35cm 20cm

Take moment about the pivot;

Clockwise moment = anticlockwise moment

W₂(20cm) = 20N(35 cm)

W₂ = (20 x 35) / (20)

W₂ = 35 N

Therefore, the weight that will balance the meter stick at 70 cm mark is 35N

8 0

2 years ago