2 years ago

Q 1-10

Physics

1 answer:

Over [174]2 years ago

6 0

The net force on the box perpendicular to the floor is

∑ F[perp] = F[normal] - mg = 0

where mg is the weight of the box. Then

F[normal] = (35.5 kg) g = 347.9 N

Since the box slides with constant speed, the net force parallel to the floor is

∑ F[para] = F[applied] - F[friction] = 0

We have

F[friction] = 0.15 F[normal] = 52.185 N

so that

F[applied] = F[friction] ≈ 52 N

You might be interested in

A close coiled helical spring of round steel wire 10 mm diameter having 10 complete turns with a mean radius of 60 mm is subject

kow [346]

Answer:

The deflection of the spring is 34.56 mm.

Explanation:

Given that,

Diameter = 10 mm

Number of turns = 10

$Radius_{mean} = 60\ mm$

$Diameter_{mean} = 120\ mm$

Load = 200 N

We need to calculate the deflection

Using formula of deflection

$\delta=\dfrac{8pD^3n}{Cd^4}$

Put the value into the formula

$\delta=\dfrac{8\times200\times(120)^3\times10}{80\times10^{3}\times10^4}$

$\delta =34.56\ mm$

Hence, The deflection of the spring is 34.56 mm.

4 0

3 years ago

Copper is used in wiring because its metallic bonding gives its a high degree of freedom and the material a high conductivity

Anettt [7]

Answer:

Yeah.

Explanation:

I guess so .

It shld be

7 0

3 years ago

A 50 kg mass is placed 2 meters from the fulcrum. To balance the lever, a second object is placed 4 meters

julia-pushkina [17]

Answer:

mass =25 kg

using clockwise moment = anticlockwise moment

8 0

3 years ago

Classify the given simple machine into lever ,pulley, wheel and axle, inclined plane, screw, wedge.

UNO [17]

Answer:

ladder: inclidine plane

broom:lever

shovel: lever

7 0

3 years ago

Based on the thermodynamic properties provided for water, determine the energy change when the temperature of 1.25 kg of water d

algol [13]

Energy change = -303050 J

Explanation:

the heat energy Q is given by the formula

Q= m C ΔT

m= mass of water=1.25 Kg

C= specific heat of water= 4180 J/Kg °C

ΔT= change in temperature =55-113= -58°C

Q= 1.25 (4180) (-58)

Q=-303050 J

Negative sign shows that the energy is released or lost.

7 0

4 years ago