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3 years ago

A 0.141 kg pinewood derby car is moving 1.33 m/s . What is its momentum?

Physics

1 answer:

kompoz [17]3 years ago

7 0

momentum= mass × velocity = 0.141kg×1.33m/s= 0.18753kg m/s = 0.188kg m/s (3s.f.)

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Starting from rest, a 2-m-long pendulum swings from an angleof

Andrews [41]

Answer:

D.) 1m/s

Explanation:

Assume the initial angle of the swing is 12.8 degree with respect to the vertical. We can calculate the vertical distance from this initial point to the lowest point by first calculate the vertical distance from this point the the pivot point:

$L_1 = L*cos(12.8) = 2*0.975 = 1.95 m$

where L is the pendulum length

The vertical distance from the lowest point to the pivot point $L_2$ is the pendulum length 2m

this means the vertical distance from this initial point to the lowest point is simply:

$L_3 = L_2 - L_1 = 2 - 1.95 = 0.05 m$

As the pendulum travel (vertically) from the initial point to the bottom point, its potential energy is converted to kinetic energy:

$E_p = E_k$

$mgh = mv^2/2$

where m is the mass of the pendulum, g = 10 m/s2 is the constant gravitational acceleration, h = 0.05 is the vertical it travels, v is the pendulum velocity at the bottom, which we are trying to solve for.

The m on both sides of the equation cancel out

$v^2 = 2gh = 2*10*0.05 = 1$

$v = \sqrt{1} = 1 m/s$

so D is the correct answer

5 0

2 years ago

Where does groundwater come from?

jek_recluse [69]

Groundwater is the water found underground in the cracks and spaces in soil, sand and rock. It is stored in and moves slowly through geologic formations of soil, sand and rocks called aquifers.

Hope this helps.

Please mark as brainliest..........

4 0

2 years ago

Read 2 more answers

While on vacation, a student picks up surface rocks from around the world to add to her rock collection. The composition of her

kykrilka [37]

I want to say that they will be primarily flat but I honestly don't know

7 0

3 years ago

3. A 2kg wooden block whose initial speed is 3 m/s slides on a smooth floor for 2 meters before it comes to a

serious [3.7K]

Answer:

Calculating Coefficient of friction is 0.229.

Force is 4.5 N that keep the block moving at a constant speed.

Explanation:

We know that speed expression is as $\mathrm{V}^{2}=\mathrm{V}_{\mathrm{i}}^{2}+2 . \mathrm{a} . \Delta \mathrm{s}$ .

Where, ${V}_{i}$ is initial speed, V is final speed, ∆s displacement and a acceleration.

Given that,

${V}_{i}$ =3 m/s, V = 0 m/s, and ∆s = 2 m

Substitute the values in the above formula,

$0=3^{2}-2 \times 2 \times a$

0 = 9 - 4a

4a = 9

$a=2.25 \mathrm{m} / \mathrm{s}^{2}$

$a=2.25 \mathrm{m} / \mathrm{s}^{2}$ is the acceleration.

Calculating Coefficient of friction:

$\mathrm{F}=\mathrm{m} \times \mathrm{a}$

$\mathrm{F}=\mu \times \mathrm{m} \times \mathrm{g}$

Compare the above equation

$\mu \times m \times g=m \times a$

Cancel "m" common term in both L.H.S and R.H.S

$\text { Equation becomes, } \mu \times g=a$

$\text { Coefficient of friction } \mu=\frac{a}{g}$

$\mathrm{g} \text { on earth surface }=9.8 \mathrm{m} / \mathrm{s}^{2}$

$\mu=\frac{2.25}{9.8}$

$\mu=0.229$

Hence coefficient of friction is 0.229.

calculating force:

$\text { We know that } \mathrm{F}=\mathrm{m} \times \mathrm{a}$

$\mathrm{F}=2 \times 2.25 \quad(\mathrm{m}=2 \mathrm{kg} \text { given })$

F = 4.5 N

Therefore, the force would be <u>4.5 N</u> to keep the block moving at a constant speed across the floor.

7 0

2 years ago

The height of a typical playground slide is about 1.8 m and it rises at an angle of 30 ∘ above the horizontal.

Salsk061 [2.6K]

Answer:

$5.94\ \text{m/s}$

$1.7$

$0.577$

Explanation:

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

$\theta$ = Angle of slope = $30^{\circ}$

v = Velocity of child at the bottom of the slide

$\mu_k$ = Coefficient of kinetic friction

$\mu_s$ = Coefficient of static friction

h = Height of slope = 1.8 m

The energy balance of the system is given by

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.8}\\\Rightarrow v=5.94\ \text{m/s}$

The speed of the child at the bottom of the slide is $5.94\ \text{m/s}$

Length of the slide is given by

$l=h\sin\theta\\\Rightarrow l=1.8\sin30^{\circ}\\\Rightarrow l=0.9\ \text{m}$

$v=\dfrac{1}{2}\times5.94\\\Rightarrow v=2.97\ \text{m/s}$

The force energy balance of the system is given by

$mgh=\dfrac{1}{2}mv^2+\mu_kmg\cos\theta l\\\Rightarrow \mu_k=\dfrac{gh-\dfrac{1}{2}v^2}{gl\cos\theta}\\\Rightarrow \mu_k=\dfrac{9.81\times 1.8-\dfrac{1}{2}\times 2.97^2}{9.81\times 0.9\cos30^{\circ}}\\\Rightarrow \mu_k=1.73$

The coefficient of kinetic friction is $1.7$ .

For static friction

$\mu_s\geq\tan30^{\circ}\\\Rightarrow \mu_s\geq0.577$

So, the minimum possible value for the coefficient of static friction is $0.577$ .

8 0

2 years ago