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2 years ago

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The amount of matter in an object

1 answer:

Mnenie [13.5K]2 years ago

6 0

Alot? You didn’t show me any work to solve

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I need help with this

TEA [102]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

One would use deductive reasoning to

Ivan

The answer is Prove a hypothesis

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3 years ago

Do the two cars ever have the same velocity at one instant of time? If so, between which two frames? Check all that apply. Check

alexgriva [62]

Answer:

Cars have the same velocity at one instant of time between dots 4 and 5.

Explanation:

after looking at the image it is visible that same distance is covered by both cars in the frame between 4 and 5

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3 years ago

Which has more momentum, a 2000 lb car moving at 100 km/hr or a 4000 lb truck moving at 50 km/hr ?

BabaBlast [244]

Answer:

The truck and car have the same momentum.

Explanation:

$m_1$ = Mass of car = $2000\ \text{lb}=2000\times0.45359237\ \text{kg}$

$v_1$ = Velocity of car = $\dfrac{100}{3.6}\ \text{m/s}$

$m_2$ = Mass of truck = $4000\times0.45359237\ \text{kg}$

$v_2$ = Velocity of truck = $\dfrac{50}{3.6}\ \text{m/s}$

Momentum of car

$p_1=m_1v_1\\\Rightarrow p_1=2000\times0.45359237\times \dfrac{100}{3.6}\\\Rightarrow p_1=25199.58\ \text{kg m/s}$

Momentum of the truck

$p_2=m_2v_2\\\Rightarrow p_2=4000\times0.45359237\times \dfrac{50}{3.6}\\\Rightarrow p_2=25199.58\ \text{kg m/s}$

Both the truck and car have the same momentum of $25199.58\ \text{kg m/s}$ .

8 0

3 years ago

A ball is moving with velocity 5 m/s in a direction which makes an angle of 30° with horizontal (i.e. with positive x-direction)

klasskru [66]

Answer:

Explanation:

(a)

From the given information:

The initial velocity $v_1$ = 5 m/s

The direction of the angle θ = 30°

Therefore, the component along the x-axis = $v_1 \ cos \ \theta$

$v_{1 \ x } = 5 \ cos \ 30^0$

$v_{1x} = 4.33 \ m/s$

The component along the y-axis = $v_2 { \ sin \ \theta}$

$v_{1 \ y } = 5 \ sin \ 30^0$

$v_{1 \ y } = 2.5 \ m/s$

To find the final velocity( reflected velocity)

using the same magnitude $v_2 = 5 \ m/s$

The angle from the x-axis can be $\theta_r = 90^0+60^0$

= 150°

Thus, the component along the x-axis = $v_2 \ cos \theta _r$

$v_{2x} = - 0.433 \ m/s$

The component along the y-axis = $v_2 \ sin \theta_r$

$v_{2y} = 5 \ sin \ 150^0$

$v_{2y} = 2.5 \ m/s$

(b)

The velocity $v_1$ can be written as in vector form.

$v_1 ^{\to} = v_1 x \hat {i} + v_1 y \hat {j}$

$v_1 ^{\to} =4.33 \ \hat {i} + 2.5 \ \hat {j}$ ---- (1)

The reflected velocity in vector form can be computed as:

$v_2 ^{\to} = v_2 x \hat {i} + v_2 y \hat {j}$

$v_2 ^{\to} =-4.33 \ \hat {i} + 2.5 \ \hat {j}$ --- (2)

The change in velocity = $v_2 ^{\to} - v_1 ^{\to}$

$\Delta v ^{\to} = - 4.33 \hat i + 2.5 \hat j - 4.33 \hat i - 2.5 \hat j$

$\Delta v ^{\to} = - 8.66 \hat { i }$

(c)

The magnitude of change in velocity = $| \Delta V |$

$| \Delta V |$ = 8.66 m/s

6 0

2 years ago