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photoshop1234 [79]

2 years ago

13

A catcher in a baseball game stops a pitched ball that was originally moving at 44 m/s over a distance of 12.5 cm. the mass of t

he ball is 0.15 kg. what is the average force that the glove imparts to the ball during the catch

1 answer:

eduard2 years ago

4 0

Answer:

Force = -1161.6 Newton

Explanation:

Given the following data;

Initial velocity, u = 44m/s

Distance ,s = 12.5cm to m = 12.5/100 = 0.125m

Mass = 0.15kg

To find the acceleration;

We would use the third equation of motion;

V ² = U² + 2as

0² = 44² + 2*a*0.125

0 = 1936 + 0.25a

0.25a = -1936

a = -1936/0.25

Acceleration, a = -7744m/s2

Force = mass * acceleration

Substituting into the equation, we have;

Force = 0.15 * (-7744)

Force = -1161.6 Newton

The value of its force is negative because the glove decreases the velocity of the ball.

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A moonshiner makes the error of filling a glass jar to the brim and capping it tightly. The moonshine expands more than the glas

ohaa [14]

Answer:

ΔP = (640 N/cm^2)

Explanation:

Given:-

- The volume increase, ΔV/V0 = 4 ✕ 10^-3

- The Bulk Modulus, B = 1.6*10^9 N/m^2

Find:-

Calculate the force exerted by the moonshine per square centimeter

Solution:-

- The bulk modulus B of a material is dependent on change in pressure or Force per unit area and change in volume by the following relationship.

B = ΔP / [(ΔV/V)]

- Now rearrange the above relation and solve for ΔP or force per unit area.

ΔP = B* [(ΔV/V)]

- Plug in the values:

ΔP = (1.6*10^9)*(4 ✕ 10^-3)

ΔP = 6400000 N/m^2

- For unit conversion from N/m^2 to N/cm^2 we have:

ΔP = (6400000 N/m^2) cm^2 / (100)^2 m^2

ΔP = (640 N/cm^2)

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3 years ago

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OlgaM077 [116]

Answer:

The speed with which the man flies forward is 5.5 m/s

Explanation:

The mass of the man = 100 kg

The mass of the scooter = 10 kg

The speed with which the man was traveling on the scooter = 5 m/s

The speed of the scooter after it hits the rock = 0 m/s

Let v represent the speed with which the man flies forward

The formula for momentum, P, is P = Mass × Velocity

The conservation of linear momentum principle is, the total initial momentum = The total final momentum, therefore, we have;

The total initial momentum = (100 kg + 10 kg) × 5 m/s = 550 kg·m/s

The total final momentum = 100 kg × v + 10 kg × 0 m/s = 100 kg × v

When the momentum is conserved, we have;

550 kg·m/s = 100 kg × v

∴ v = 550 kg·m/s/(100 kg) = 5.5 m/s.

The speed with which the man flies forward = v = 5.5 m/s

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2 years ago

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mrs_skeptik [129]

Explanation:

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2 years ago

Read 2 more answers

A 44-cm-diameter water tank is filled with 35 cm of water. A 3.0-mm-diameter spigot at the very bottom of the tank is opened and

cricket20 [7]

Answer:

The frequency $f$ = 521.59 Hz

The rate at which the frequency is changing = 186.9 Hz/s

Explanation:

Given that :

Diameter of the tank = 44 cm

Radius of the tank = $\frac{d}{2}$ = $\frac{44}{2}$ = 22 cm

Diameter of the spigot = 3.0 mm

Radius of the spigot = $\frac{d}{2}$ = $\frac{3.0}{2}$ = 1.5 mm

Diameter of the cylinder = 2.0 cm

Radius of the cylinder = $\frac{d}{2}$ = $\frac{2.0}{2}$ = 1.0 cm

Height of the cylinder = 40 cm = 0.40 m

The height of the water in the tank from the spigot = 35 cm = 0.35 m

Velocity at the top of the tank = 0 m/s

From the question given, we need to consider that the question talks about movement of fluid through an open-closed pipe; as such it obeys Bernoulli's Equation and the constant discharge condition.

The expression for Bernoulli's Equation is as follows:

$P_1+\frac{1}{2}pv_1^2+pgy_1=P_2+\frac{1}{2}pv^2_2+pgy_2$

$pgy_1=\frac{1}{2}pv^2_2 +pgy_2$

$v_2=\sqrt{2g(y_1-y_2)}$

where;

P₁ and P₂ = initial and final pressure.

v₁ and v₂ = initial and final fluid velocity

y₁ and y₂ = initial and final height

p = density

g = acceleration due to gravity

So, from our given parameters; let's replace

v₁ = 0 m/s ; y₁ = 0.35 m ; y₂ = 0 m ; g = 9.8 m/s²

∴ we have:

v₂ = $\sqrt{2*9.8*(0.35-0)}$

v₂ = $\sqrt {6.86}$

v₂ = 2.61916

v₂ ≅ 2.62 m/s

Similarly, using the expression of the continuity for water flowing through the spigot into the cylinder; we have:

v₂A₂ = v₃A₃

v₂r₂² = v₃r₃²

where;

v₂r₂ = velocity of the fluid and radius at the spigot

v₃r₃ = velocity of the fluid and radius at the cylinder

$v_3 = \frac{v_2r_2^2}{v_3^2}$

where;

v₂ = 2.62 m/s

r₂ = 1.5 mm

r₃ = 1.0 cm

we have;

v₃ = $(2.62 m/s)*$ $(\frac{1.5mm^2}{1.0mm^2} )$

v₃ = 0.0589 m/s

∴ velocity of the fluid in the cylinder = 0.0589 m/s

So, in an open-closed system we are dealing with; the frequency can be calculated by using the expression;

$f=\frac{v_s}{4(h-v_3t)}$

where;

$v_s$ = velocity of sound

h = height of the fluid

v₃ = velocity of the fluid in the cylinder

$f=\frac{343}{4(0.40-(0.0589)(0.4)}$

$f= \frac{343}{0.6576}$

$f$ = 521.59 Hz

∴ The frequency $f$ = 521.59 Hz

b)

What are the rate at which the frequency is changing (Hz/s) when the cylinder has been filling for 4.0 s?

The rate at which the frequency is changing is related to the function of time (t) and as such:

$\frac{df}{dt}= \frac{d}{dt}(\frac{v_s}{4}(h-v_3t)^{-1})$

$\frac{df}{dt}= -\frac{v_s}{4}(h-v_3t)^2(-v_3)$

$\frac{df}{dt}= \frac{v_sv_3}{4(h-v_3t)^2}$

where;

$v_s$ (velocity of sound) = 343 m/s

v₃ (velocity of the fluid in the cylinder) = 0.0589 m/s

h (height of the cylinder) = 0.40 m

t (time) = 4.0 s

Substituting our values; we have ;

$\frac{df}{dt}= \frac{343*0.0589}{4(0.4-(0.0589*4.0))^2}$

= 186.873

≅ 186.9 Hz/s

∴ The rate at which the frequency is changing = 186.9 Hz/s when the cylinder has been filling for 4.0 s.

8 0

3 years ago

A family is skating at an ice rink. The 58.2 kg mother is holding the

MariettaO [177]

Answer:

When I got this question I had to draw it out so if you have to do that, draw 3 stick figures holding hands, one representing the mother, father, and daughter. Then you write their weights on top of them and then draw an arrow pointing from the father to the mother.

Explanation:

use this formula :

$a_{y}$ = $\frac{Fdadshandy}{msys}$

then you fill it in :

$a_{y}$ = $\frac{100N}{35.5kg+58.2kg}$

$a_{y}$ = $\frac{100N}{93.7kg}$

$a_{y}$ = $1.0672$ $m/s^{2}$

then you multiply that with the daughters weight :

$T_{2} x= m_{2} a_{y}$

$T_{2} x = 35.5kg (1.0672 m/s^{2})$

$T_{2} x = 37.89N$

and that's the answer :) : 37.89N

5 0

3 years ago