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3 years ago

A constant force of 8.0 N is exerted for 4.0 s on a 16-kg object initially at rest. The change in speed of this object will be:

Physics

1 answer:

AfilCa [17]3 years ago

8 0

The change in speed of this object is 3m/s

According to Newton's second law;

F = ma

F = mv/t

Given the following parameters

Force F = 8.0N

mass m = 16kg

time t = 4.0s

Required

speed v

Substitute the given parameters into the formula

v = Ft/m

v = 8 * 6/16

v = 48/16

v = 3m/s

Hence the change in speed of this object is 3m/s

Learn more here: brainly.com/question/19072061

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Thank your very much

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3 years ago

A 3 mm inside diameter tube is placed in a fluid with a surface tension of 600 mN/m and density of 3.7 g/cm3. The contact angle

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Answer: The height of the fluid rise is 0.01m

Explanation:

Using the equation

h = (2TcosѲ )/rpg

h= height of the fluid rise

diameter of the tube =3mm

radius of the tube= 3/2 =1.5mm=0.0015

T= surface tension = 600mN/m=0.6N/m

Ѳ = contact angle = $60^o$ C

p= density =3.7g/cm3= 3700kg/m3

g= acceleration due to gravity =9.8m/s2

h = ( 2*0.6*0.5)/(0.0015*3700*9.8)

h = 0.6/54.39

h= 0.01m

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3 years ago

An electron is released from rest in a uniform electric field and accelerates to the north at a rate of 115 m/s squared. What ar

Iteru [2.4K]

The direction of the electric field would be south.

qE/m = 115
 E = 115*m/q 
 = 115 * 9.1 * 10^(-31) / 1.67*10^(-19) 
 = 762.87 * 10^(-12) 
 = 6.27 x 10^-10 N/C

Hope this answers the question. Have a nice day. Feel free to ask more questions.

6 0

3 years ago

Tammy leaves the office, drives 34 km due north, then turns onto a second highway and continues in a direction of 35◦ north of e

Natasha2012 [34]

Answer:

The total displacement is 102 km $51^o$ north of east.

Explanation:

We can treat this problem as a trigonometric one, so we need to calculate the total displacement on the north and east.

$d_n=34km+79km*sin(35^o)=79km$

and

$d_e=79*cos(35^o)=65km$

The total displacement is given by:

$D=\sqrt{(79km)^2+(65km)^2}=102km$

with an agle of:

$\alpha =arctg(\frac{79km}{65km})=51^o$

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3 years ago

The elements least likely to form bonds are found in which group? O 15 O 16 017 O 18

12345 [234]

The answer is group 18.

Explanation:
Group 18 is known as the Noble Gases. They are the most stable elements because they have a full outer shell of valence electrons, so they have no need to bond with other elements.

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3 years ago

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