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tekilochka [14]

3 years ago

11

Does gravity exist between a pencil on a coffee mug

1 answer:

DIA [1.3K]3 years ago

5 0

Answer:

Yes it does

Explanation:

Gravity is pushing down on the pencil but the coffee mug is also pushing the pencil up with the same amount of force so they both don't move

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The drag on a pitched baseball can be surprisingly large. Suppose a 145 g baseball with a diameter of 7.4 cm has an initial spee

kupik [55]

Answer:

<h2>Part A)</h2><h2>Acceleration of the ball is 10.1 m/s/s</h2><h2>Part B)</h2><h2>the final speed of the ball is given as</h2><h2> $v_f = 35.3 m/s$ </h2>

Explanation:

Part a)

As we know that drag force is given as

$F = \frac{C_d \rho A v^2}{2}$

$C_d = 0.35$

$A = \frac{\pi d^2}{4}$

$A = \frac{\pi(0.074)^2}{4}$

$A = 4.3 \times 10^{-3} m^2$

$v = 40.2 m/s$

so we have

$F = \frac{0.35\times 1.2 (4.3 \times 10^{-3})(40.2)^2}{2}$

$F = 1.46 N$

So acceleration of the ball is

$a = \frac{F}{m}$

$a = \frac{1.46}{0.145}$

$a = 10.1 m/s^2$

Part B)

As per kinematics we know that

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 40.2^2 = 2(-10.1)(18.4)$

$v_f = 35.3 m/s$

4 0

3 years ago

The steps taken in scientific studies often follow similar patterns. which answer correctly describes the order in which the ste

scZoUnD [109]

OPTION A WOULD BE YOUR ANSWER

7 0

3 years ago

Read 2 more answers

If I could lift up to ten tons and I threw a ball the size of an orange but weighed a ton, to the ground, how big of an impact w

Anarel [89]

Answer:

The impact force is 98000 N.

Explanation:

mass = 10 tons

The impact force is the weight of the object.

Weight =mass x gravity

W = 10 x 1000 x 9.8

W = 98000 N

The impact force is 98000 N.

5 0

3 years ago

Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit

Ede4ka [16]

Answer:

$386.2^{\circ}F$

Explanation:

We are given that

$P_1=200lbf/in^2$

$P_2=60lbf/in^2$

$v_1=200ft/s$

$v_2=1700ft/s$

$T_1=500^{\circ}F$

$Q=0$

$C_p=1BTU/lb^{\circ}F$

We have to find the exit temperature.

By steady energy flow equation

$h_1+v^2_1+Q=h_2+v^2_2$

$C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}$

$1BTU/lb=25037ft^2/s^2$

Substitute the values

$1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}$

$500+1.598=T_2+115.4$

$T_2=500+1.598-115.4$

$T_2=386.2^{\circ}F$

7 0

3 years ago

A 6 kg box with initial speed 5 m/s slides across the floor and comes to a stop after 1.9 s. What is the coefficient of kinetic

Ilia_Sergeevich [38]

Answer:

$\mu_k=0.27$

Explanation:

According to the free body diagram, in this case, we have:

$\sum F_x:-F_k=ma\\\sum F_y:N=mg$

Recall that the force of friction is given by:

$F_k=\mu_k N$

Replacing and solving for the coefficient of kinetic friction:

$-\mu_kN=ma\\-\mu_k(mg)=ma\\\mu_k=-\frac{a}{g}$

We have an uniformly accelerated motion. Thus, the acceleration is defined as:

$a=\frac{v_f-v_0}{t}\\a=\frac{0-5\frac{m}{s}}{1.9s}\\a=-2.63\frac{m}{s^2}$

Finally, we calculate $\mu_k$ :

$\mu_k=-\frac{-2.63\frac{m}{s^2}}{9.8\frac{m}{s^2}}\\\mu_k=0.27$

4 0

3 years ago