Ask question

Ask question

All categories

anyanavicka [17]

3 years ago

7

A 97.6-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk =

0.555.
(a) What is the magnitude of the frictional force?
(b) If the player comes to rest after 1.22 s, what is his initial speed?

1 answer:

Mila [183]3 years ago

4 0

Answer:

$v=6.65m/sec$

Explanation:

From the Question we are told that:

Mass $m=97.6$

Coefficient of kinetic friction $\mu k=0.555$

Generally the equation for Frictional force is mathematically given by

$F=\mu mg$

$F=0.555*97.6*9.8$

$F=531.388N$

Generally the Newton's equation for Acceleration due to Friction force is mathematically given by

$a_f=-\mu g$

$a_f=-0.555 *9.81$

$a_f=-54455m/sec^2$

Therefore

$v=u-at$

$v=0+5.45*1.22$

$v=6.65m/sec$

You might be interested in

MA of the first class lever may be equal to, greater than 1.Why

Anika [276]

Answer:

hjhjhjhjhjhj

Explanation:

jmjk

4 0

3 years ago

At one atmosphere of pressure and 25°C, a hot air balloon has a volume of 4,000 liters. While still tied to the ground, the air

yawa3891 [41]

Based on the options given, the most likely answer to this query is C) 4577 liters.

Upon computation of the given variables the result seems to be 4577 L

Thank you for your question. Please don't hesitate to ask in Brainly your queries.

3 0

3 years ago

Read 2 more answers

A security review has flagged this architecture as vulnerable, and a Security Engineer has been asked to make this design more s

dybincka [34]

Answer:

Complete Question:

A company has two AWS accounts, each containing one VPC. The first VPC has a VPN connection with its corporate network. The second VPC, without a VPN, hosts an Amazon Aurora database cluster in private subnets. Developers manage the Aurora database from a bastion host in a public subnet as shown in the image.

A security review has flagged this architecture as vulnerable, and a Security Engineer has been asked to make this design more secure. The company has a short deadline and a second VPN connection to the Aurora account is not possible.

How can a Security Engineer securely set up the bastion host?

A. Move the bastion host to the VPC and VPN connectivity. Create a VPC peering relationship between the bastion host VPC and Aurora VPC.

B. Create a SSH port forwarding tunnel on the Developer's workstation to the bastion host to ensure that only authorized SSH clients can access the bastion host.

C. Move the bastion host to the VPC with VPN connectivity. Create a cross-account trust relationship between the bastion VPC and Aurora VPC, and update the Aurora security group for the relationship.

D. Create an AWS Direct Connect connection between the corporate network and the Aurora account, and adjust the Aurora security group for this connection.

Answer:

B. Create an SSH port forwarding tunnel on the Developer's workstation to the bastion host to ensure that only authorized SSH clients can access the bastion host.

Explanation:

To gain a better understanding of why the option selected in the answer to the question let first explain some terms.

AWS:

According to techtarget,

AWS (Amazon Web Services) is a comprehensive, evolving cloud computing platform provided by Amazon that includes a mixture of (1) infrastructure as a service (IaaS),(2) platform as a service (PaaS) and (3)packaged software as a service (SaaS) offerings.

An AWS account is a container for your AWS resources

A bastion host is a server whose purpose is to provide access to a private network from an external network, such as the Internet. Because of its exposure to potential attacks, a bastion host must minimize the chances of penetration to the private network.

SSH port forwarding, or TCP/IP connection tunneling, is a process whereby a TCP/IP connection that would otherwise be insecure is tunneled through a secure SSH(Secure Shell (SSH) is a cryptographic network protocol for operating network services securely over an unsecured network.) link, thus protecting the tunneled connection from network attacks.

So the Bastion protects the private network while the SSH prevent unauthorized access to the bastion

6 0

3 years ago

yKpoI14uk [10]

13.6g

136g

Explanation:

Given parameters:

Density of mercury = 13.6g/cm³

Unknown:

Mass of :

1cm³ of mercury

10cm³ of mercury

Solution:

Density is the mass per unit volume of a substance. It is the amount of matter in a given volume.

Density = $\frac{mass}{volume}$

Since the unknown in the problem is mass;

Mass = density x volume

Mass of 1cm³ of mercury = 13.6 x 1 = 13.6g

Mass of 10cm³ of mercury = 10 x 13.6 = 136g

learn more:

Density problems brainly.com/question/3433940

#learnwithBrainly

5 0

3 years ago

The basic barometer can be used to measure the height of a building. If the barometric readings at the top and at the bottom of

Elan Coil [88]

Answer: 230.50 m

Explanation:

We have the following information:

$h_{Hg-TOP}=675mmHg=0.675m$ the barometric reading at the top of the building

$h_{Hg-BOT}=695mmHg=0.695m$ the barometric reading at the bottom of the building

$\rho _{air}=1.18 kg/m^{3}$ density of air

$\rho _{Hg}=13600 kg/m^{3}$ density of mercury

$g=9.8/m^{2}$ gravity

And we need to find the height of the building.

In order to approach this problem, we will firstly use the following equations to find the pressure at the top of the building $P_{TOP}$ and the perssure at the bottom $P_{BOT}$ :

$P_{TOP}=\rho _{Hg} g h_{Hg-TOP}$ (1)

$P_{BOT}=\rho _{Hg} g h_{Hg-BOT}$ (2)

From (1): $P_{TOP}=(13600 kg/m^{3})(9.8/m^{2})(0.675m)=89964 Pa$ (3)

From (2): $P_{BOT}=(13600 kg/m^{3})(9.8/m^{2})(0.695m)=92629.6 Pa$ (4)

Having the pressures at the top and the bottom of the building, we can calculate the variation in pressure $\Delta P$ :

$\Delta P=P_{BOT} - P_{TOP}$ (5)

$\Delta P=92629.6 Pa - 89964 Pa=2665.6 Pa$ (6)

On the other hand, we have a column of air with a cross-section area $A$ and the same height of the building, lets name it $h_{air}$ .

As pressure is defined as the force $F$ exerted on a specific area $A$ , we can write:

$\Delta P=\frac{F}{A}$ (7)

If we isolate $F$ we have:

$F= A \Delta P$ (8)

Also, the force gravity exerts on this column of air (its weight) is:

$F=m_{air} g$ (9)

Knowing the density of air is: $\rho_{air}=\frac{m_{air}}{V_{air}}$ (10)

where the volume of air can be written as: $V_{air}=(A)(h_{air})$ (11)

Substituting (1) in (10):

$\rho_{air}=\frac{m_{air}}{(A)(h_{air}}$ (12)

Isolating $m_{air}$ :

$m_{air}=(\rho_{air}) (A) (h_{air})$ (13)

Substituting (13) in (9):

$F=(\rho_{air}) (A) (h_{air}) (g)$ (14)

Matching (8) and (14)

$A \Delta P=(\rho_{air}) (A) (h_{air}) (g)$ (15)

Isolating $h_{air}$ :

$h_{air}=\frac{\Delta P}{g \rho_{air}}$ (16)

Substituting the known and calculated values:

$h_{air}=\frac{2665.6 Pa}{(9.8m/s^{2}) (1.18 kg/m^{3})}$ (17)

Finally:

$h_{air}=230.50 m$ This is the height of the building

8 0

3 years ago