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garik1379 [7]
4 years ago
10

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

rock samples will be primarily
Physics
1 answer:
kykrilka [37]4 years ago
7 0
I want to say that they will be primarily flat but I honestly don't know
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What is the kinetic energy of a 2995 kg truck when it moving 47 ms
Dafna11 [192]
To find the kinetic energy . You have to use this equation : 1/2m(v^2) .

7 0
3 years ago
A car that increases its speed from 20 km/h to 100 km/h undergoes
Kamila [148]

Answer:

accelaration

Explanation:

3 0
3 years ago
A basketball with mass of 0.8 kg is moving to the right with velocity 6 m/s and hits a volleyball with mass of 0.6 kg that stays
IceJOKER [234]

Answer:

26.67 m/s

Explanation:

From the law of conservation of linear momentum, the initial sum of momentum equals the final sum.

p=mv where p is momentum, m is the mass of object and v is the speed of the object

Initial momentum

The initial momentum will be that of basketball and volleyball, Since basketball is initially at rest, its initial velocity is zero

p_i= m_bv_b+m_vv_v=8*6+0.6*0=48 Kg.m/s

Final momentum

p_f= m_bv_b+m_vv_v=8*4+0.6*v_v=32+0.6v Kg.m/s\\32+0.6v_v=48\\0.6v=16\\v_v=16/0.6=26.66666667\approx 26.67 m/s

4 0
3 years ago
A block of mass M on a horizontal surface is connected to the end of a massless spring of spring constant k. The block is pulled
slamgirl [31]

Answer:

Minimum coefficient of kinetic friction between the surface and the block is \mu_k=\frac{kx}{2Mg} .

Explanation:

Given:

Mass of the block = M

Spring constant = k

Distance pulled = x

According to the question:

<em>We have to find the minimum co-efficient of kinetic friction between the surface and the block that will prevent the block from returning to its equilibrium with non-zero speed.  </em>

So,

From the FBD we can say that:

⇒ Normal force, N=Mg                                   <em>...equation(i)</em>

⇒ Elastic potential energy, PE = \frac{kx^2}{2}               <em>  ...equation (ii)</em>

⇒ Frictional force, f = \mu_kN                                <em> ...equation (iii)</em>

⇒ Plugging (i) in (iii).

⇒ f=\mu_kMg

Now,

⇒ As we know that the energy lost due to friction is equivalent to PE .

⇒ PE=fx                     <em>...considering PE as</em> mgh or f(x) .

   Arranging the equation.

⇒ \frac{kx^2}{2}=\mu_k Mg (x)

⇒ \frac{kx}{2}=\mu_k Mg                 <em>...eliminating x from both sides.</em>

⇒ \frac{kx}{2Mg}=\mu_k                    <em>...dividing both sides wit Mg.</em>

Minimum coefficient of kinetic friction between the surface and the block is \frac{kx}{2Mg}=\mu_k .

4 0
4 years ago
You throw a rock straight up from the edge of a cliff. It leaves your hand at time t = 0 moving at 13.0 m/s. Air resistance can
Zarrin [17]

Answer:

0.36s, 2.3s

Explanation:

Let gravitational acceleration g = 9.81 m/s2. And let the throwing point as the ground 0 for the upward motion. The equation of motion for the rock leaving your hand can be written as the following:

s = v_0t + gt^2/2

where s = 4 m is the position at 4m above your hand. v_0 = 13 m/s is the initial speed of the rock when it leaves your hand. g = -9.81m/s2 is the deceleration because it's in the downward direction. And t it the time(s) it take to get to 4m, which we are looking for

4 = 13t - 9.81t^2/2

4.905 t^2 - 13t + 4 = 0

t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

t= \frac{13\pm \sqrt{(-13)^2 - 4*(4.905)*(4)}}{2*(4.905)}

t= \frac{13\pm9.51}{9.81}

t = 2.3 or t = 0.36

7 0
4 years ago
Read 2 more answers
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