Ask question

Ask question

All categories

2 years ago

9

A pitcher throws a 0.15 kg baseball so that it crosses home plate horizontally with a speed of 10 m/s. It is hit straight back a

t the pitcher with a final speed of 24 m/s. Assume the direction of the initial motion of the baseball to be positive.

1 answer:

Maru [420]2 years ago

4 0

Answer:

-5.1 kg m/s

Explanation:

Impulse is the change in momentum.

Change in momentum= final momentum - initial momentum=m $v_{2}$ +m $v_{1}$

Plugging in the values= -0.15*24 - (0.15*10) (The motion towards the pitcher is negative as the initial motion is considered to be positive)

Impulse=-5.1 kg m/s (-ve means that it is the impulse towards the pitcher)

You might be interested in

MGS and later orbiters found spectral evidence for what minerals on the Martian surface? Check all that apply.

slamgirl [31]

Mars Global Surveyors (MGS) and later orbiters found the following minerals on the Martian surface;

Phyllosilicates

Carbonate
Sulfates
Iron oxide

The Mars Global Surveyors (MGS) and later orbiters suggest that the Martian crust contains a higher percentage of volatile elements such as Sulphur and chlorine than the Earth's crust does.

These scientists also conclude that the most abundant chemical elements in the Martian crust are those found in Igneous rock.

These elements include the following;

Silicon,
Oxygen,
Iron,
Magnesium,
Aluminum,
Calcium, and
Potassium.

They also, suggest that hydrogen is found in ice (water) while carbon is found in carbon dioxide and carbonates.

From the given options the minerals found in Martian surface include;

Phyllosilicates ------ these are sheet of silicate minerals

Carbonate
Sulfates
iron oxide

Learn more here: brainly.com/question/20470323

6 0

2 years ago

One of the characteristic of sound enogy Similar<br>to light energy is?

Nastasia [14]

Answer:

They both produce heat energy.

6 0

2 years ago

you have been called to testify as a as an expert witness in a trial involving a head-on collision Car A weighs 1515 pounds and

GrogVix [38]

Answer:

70.6 mph

Explanation:

Car A mass= 1515 lb

Car B mass=1125 lb

Speed of car B is 46 miles/h

Distance before locking, d=19.5 ft

Coefficient of kinetic friction is 0.75

Initial momentum of car B=mv where m is mass and v is velocity in ft/s

46 mph*1.46667=67.4666668 ft/s

$Momentum_B=1125*67.4666668 ft/s$

Initial momentum of car A is given by

$Momentum_A=1515v_a$ where $v_a$ is velocity of A

Taking East as positive and west as negative then the sum of initial momentum is

$1515v_a-(1125*67.4666668 ft/s)$

The common velocity is represented as $v_c$ hence after collision, the final momentum is

$Momentum_final=(m_a+m_b)v_c=(1515+1125)v_c=2640v_c$

From the law of conservation of linear momentum, sum of initial and final momentum equals each other hence

$1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

The acceleration of two cars $a=-\mug=-0.75*32.17=-24.1275 ft/s^{2}$

From kinematic equation

$v^{2}=u^{2}+2as$ hence

$v^{2}-u^{2}=2as$

$0^{2}-(v_c)^{2}=2*-24.1275*19.5$

$v_c=\sqrt{2*24.1275*19.5}=30.67 ft/s$

Substituting the value of $v_c$ in equation $1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

$1515v_a-(1125*67.4666668 ft/s)= 2640*30.67$

$1515v_a=(1125*67.4666668 ft/s)+2640*30.67$

$v_a=\frac {156868.8}{1515}=103.5438 ft/s$

$\frac {103.5438}{1.46667}=70.59787 mph\approx 70.60 mph$

3 0

2 years ago

A light ray incident from medium 1 to medium 2, where n1>n2. When the incident angle exceed the critical angle ac, the refrac

vovikov84 [41]

Explanation:

(a)

Critical angle is the angle at the angle of refraction is 90°. After the critical angle, no refraction takes place.

Using Snell's law as:

$n_1\times {sin\theta_i}={n_2}\times{sin\theta_r}$

Where,

${\theta_i}$ is the angle of incidence

${\theta_r}$ is the angle of refraction = 90°

${n_2}$ is the refractive index of the refraction medium

${n_1}$ is the refractive index of the incidence medium

Thus,

$n_1\times {sin\ \theta_{critical}}={n_2}\times{sin\ 90^0}$

The formula for the calculation of critical angle is:

${sin\theta_{critical}}=\frac {n_2}{n_1}$

Where,

${\theta_{critical}}$ is the critical angle

(b)

No it cannot occur. It only occur when the light ray bends away from the normal which means that when it travels from denser to rarer medium.

7 0

3 years ago

What is the electric potential, i.e. the voltage, 0.30 m from a point charge of 6.4 x 10-C?

Gwar [14]

Answer:

V = 192 kV

Explanation:

Given that,

Charge, $q=6.4\times 10^{-6}\ C$

Distance, r = 0.3 m

We need to find the electric potential at a distance of 0.3 m from a point charge. The formula for electric potential is given by :

$V=\dfrac{kq}{r}\\\\V=\dfrac{9\times 10^9\times 6.4\times 10^{-6}}{0.3}\\\\V=192000\ V\\\\V=192\ kV$

So, the required electric potential is 192 kV.

3 0

2 years ago