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4 years ago

6

Which distance–time graph most closely represents an object moving with uniform motion? and for the second one, Which speed–time

graph most closely represents an object travelling with uniform motion?

1 answer:

Georgia [21]4 years ago

7 0

Answer:

1. B

2. D

Explanation:

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Write any three utilities of plastics

Artyom0805 [142]

Explanation:

container, bottles, boxes, cups

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3 years ago

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4 years ago

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The Problems: 1. Xavier starts at a position of 0 m and moves with an average speed of 0.50 m/s for 3.0 seconds. He normally mov

NemiM [27]

Answer:

(1). His final position is 1.5 m.

(2). The final position of the hedgehog is 3 m.

(3). The final position of the tortoise

(4). Her race time is 80 sec.

(5). It take to finish in 5 hr.

Explanation:

(1). Given that,

Initial position = 0 m

Average speed = 0.50 m/s

Time = 3.0 s

We need to calculate the final position

Using formula of average speed

$v_{av}=\dfrac{x_{f}+x_{i}}{t}$

Where, $x_{f}$ = final position

$x_{i}$ = Initial position

t = total time

Put the value into the formula

$0.50=\dfrac{x_{f}+0}{3.0}$

$x_{f}=0.50\times3.0$

$x_{f}=1.5\ m$

(2). Given that,

Initial position = 0 m

Average speed = 0.75 m/s

Time = 4.0 s

We need to calculate the final position

Using formula of average speed

$v_{av}=\dfrac{x_{f}+x_{i}}{t}$

Put the value into the formula

$0.75=\dfrac{x_{f}+0}{4.0}$

$x_{f}=0.75\times4.0$

$x_{f}=3\ m$

(3). Given that,

Average speed = 1.25 m/s

Time = 3.0 sec

Initial position = 1.0 m

We need to calculate the final position

Using formula of average speed

$v=\dfrac{x_{f}+x_{i}}{t}$

Put the value into the formula

$1.25=\dfrac{x_{f}+1.0}{3.0}$

$x_{f}=1.25\times3.0-1.0$

$x_{f}=2.75\ m$

(4). Given that,

Average speed = 1.25 m/s

Distance = 100 m

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v}$

Put the value into the formula

$t=\dfrac{100}{1.25}$

$t=80 sec$

(5). Given that,

Average speed = 5 miles/hr

Suppose, distance = 25 miles

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v}$

Put the value into the formula

$t=\dfrac{25}{5}$

$t=5\ hr$

Hence, (1). His final position is 1.5 m.

(2). The final position of the hedgehog is 3 m.

(3). The final position of the tortoise

(4). Her race time is 80 sec.

(5). It take to finish in 5 hr.

5 0

4 years ago

How is it possible for two objects to have the same momentum but different velocities give an example

castortr0y [4]

Momentum is
p = mv, where m is the mass and v is the speed.
So, if two objects have the same momentum but different speeds, the must have different masses as well.
v1/v2 = m2/m1
The example is a car and a truck. Car has the mass 1000 kg and velocity 100 m/s, and truck has the mass 5000 kg and velocity 20 m/s. Use the formula above to determine is it true that they have the same momentum:
(100 m/s) / (20 m/s) = (5000 kg) / (1000 kg)
5 = 5
So, the example is correct.
Hope this helps! Good luck!

4 0

3 years ago

A particle passes through the point P=(−3,1,0)P=(−3,1,0) at time t=3t=3, moving with constant velocity v⃗ =⟨5,3,−2⟩v→=⟨5,3,−2⟩.

eimsori [14]

Answer:

The parametric equation for the position of the particle is $(-18+5t,-8+3t,6-2t)$ .

Explanation:

Given that,

The point is

$P=(-3,1,0)$

Time t = 3

Velocity $v=(5,3,-2)$

We need to calculate the parametric equation for the position of the particle

Using parametric equation for position

$r(t)=r_{0}+v(t)t$ ....(I)

at t = 3,

$P=r(t)$

Put the value into the formula

$(-3,1,0)=r_{0}+(5,3,-2)\times3$

$(-3,1,0)=r_{0}+(15,9,-6)$

$r_{0}=(-18,-8,6)$

Put the value of r₀ in equation (I)

$r(t)=(-18,-8,6)+(5,3,-2)t$

$r(t)=(-18+5t,-8+3t,6-2t)$

Hence, The parametric equation for the position of the particle is $(-18+5t,-8+3t,6-2t)$ .

3 0

3 years ago