Ask question

Ask question

All categories

3 years ago

11

A 0.45kg baseball is pitched towards home plate at 20 m/s. The ball is hit back towards the pitcher with a speed of 30 m/s. What

is the change in momentum of the baseball?

1 answer:

Inessa05 [86]3 years ago

3 0

Answer:

4.5kgm/s

Explanation:

Change in momentum is expressed as

Change in momentum = m(v-u)

M is the mass

V is the final velocity

u is the initial velocity

Given

m=0.45kg

v = 30m/s

u = 20m/s

Substitute

Change in momentum = 0.45(30-20)

Change in momentum = 0.45×10

Change in momentum = 4.5kgm/s

You might be interested in

At which temperature does the motion of atoms and molecules stop?<br> 0°C<br> 0C<br> 0°K<br> 0K

anastassius [24]

Answer: 0K

Explanation:

Absolute 0 (0K) is the point where nothing could be colder and no heat energy remains in a substance.

7 0

3 years ago

What are 3 ways a car can accelerate? (CRE)

Nookie1986 [14]

Answer - There are three ways an object can accelerate: a change in velocity, a change in direction, or a change in both velocity and direction.

6 0

3 years ago

Read 2 more answers

a man is standing near the edge of a cliff 85 meters high. he throws a stone upward vertically with an intial velocity of 10 m/s

alex41 [277]

Answer:

h = 90.10 m

Explanation:

Given that,

A man is standing near the edge of a cliff 85 meters high, h₀ = 85 m

The initial speed of the stone, u = 10 m/s

The path followed by the projectile is given by :

$h(t)=-4.9t^2+10t+85$ ....(1)

For maximum height,

Put dh/dt = 0

So,

$\dfrac{dh}{dt}=-9.8t+10=0\\\\t=\dfrac{10}{9.8}\\\\=1.02\ s$

Put the value of t in equation (1).

$h(t)=-4.9(1.02)^2+10(1.02)+85\\\\=90.10\ m$

So, the maximum height of the stone is equal to 90.10 m.

5 0

3 years ago

One end of an insulated metal rod is maintained at 100 ∘C and the other end is maintained at 0.00 ∘C by an ice–water mixture. Th

lozanna [386]

Answer:

$k=105.0359\times 10^4\,W.m^{-1}.K^{-1}$

Explanation:

Given:

temperature at the hotter end, $T_H=100^{\circ}C$

temperature at the cooler end, $T_C=0^{\circ}C$

length of rod through which the heat travels, $dx=0.7\,m$

cross-sectional area of rod, $A=1.1\times 10^{-4}\,cm^2$

mass of ice melted at zero degree Celsius, $m=8.7\times 10^{-3}\,kg$

time taken for the melting of ice, $t=15\times60=900\,s$

thermal conductivity k=?

By Fourier's Law of conduction we have:

$\dot{Q}=k.A.\frac{dT}{dx}$ ......................................(1)

where:

$\dot{Q}$ =rate of heat transfer

dT= temperature difference across the length dx

Now, we need the total heat transfer according to the condition:

we know the latent heat of fusion of ice, $L = 334000\,J.kg^{-1}$

$Q=m.L$

$Q=8.7\times 10^{-3}\times 334000$

$Q=2905.8\,J$

Now the heat rate:

$\dot{Q}=\frac{Q}{t}$

$\dot{Q}=\frac{2905.8}{900}$

$\dot{Q}=3.2287\,W$

Now using eq,(1)

$3.2287=k\times 1.1\times 10^{-4} \times \frac{100}{0.7}$

$k=205.4606\,W.m^{-1}.K^{-1}$

8 0

3 years ago

if vector u has lenght 70 and direction 40 degrees, and vector v has length 85 and direction 335 degrees what is the length and

Anastaziya [24]

Answer:

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

Explanation:

||u|| = 70

θ = 40°

$\vec{u}_x=||u||cos\theta \\\Rightarrow \vec{u}_x=70cos40=53.62$

$\vec{u}_y=||u||sin\theta \\\Rightarrow \vec{u}_y=70sin40=44.99$

||v|| = 85

θ = 335°

$\vec{v}_x=||v||cos\theta \\\Rightarrow \vec{v}_x=85cos335=77.03$

$\vec{v}_y=||v||sin\theta \\\Rightarrow \vec{v}_y=85sin335=-35.92$

Resultant

$R=\sqrt{R_x^2+R_y^2}\\\Rightarrow R=\sqrt{(\vec{u}_x+\vec{v}_x)^2+(\vec{u}_y+\vec{v}_y)^2}\\\Rightarrow R =\sqrt{(70cos40+85cos335)^2+(70sin40+85sin335)^2}\\\Rightarrow R =131.15$

$\theta=tan^{-1}\frac{R_y}{R_x}\\\Rightarrow \theta=tan^{-1}\frac{70sin40+85sin335}{70cos40+85cos335}\\\Rightarrow \theta=tan^{-1}0.069=3.97^{\circ}$

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

4 0

3 years ago