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2 years ago

Larry lightweight stands on a pair of bathroom scales each scale reads 300 N what is larry's mass

Physics

2 answers:

leonid [27]2 years ago

7 0

Larry's mass is <span>30.6 kg.</span>

Luda [366]2 years ago

6 0

Answer:

Mass = 30.6 kg

Explanation:

Given

Weight scale reading W = 300 N

Solution

Weight W = Mass (m) x Acceleration due to gravity (g)

W = mg

m = W/g

m = 300 / 9.81

m = 30.6 Kg

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A block of wood has density 0.500 g/cm3 and mass 2 000 g. It floats in a container of oil (the oil's density is 0.750 g/cm3). Wh

Gennadij [26K]

Answer:

2,666.67cm^3

Explanation:

All we need to do in this problem is divide the mass of the wooden block to the oil's density.

2000/0.750 ≈ 2,666.67cm^3

Best of Luck!

6 0

1 year ago

A 1000-kg car is slowly picking up speed as it goes around a horizontal curve whose radius is 100 m. The coefficient of static f

Snezhnost [94]

Answer:

18.5 m/s

Explanation:

On a horizontal curve, the frictional force provides the centripetal force that keeps the car in circular motion:

$\mu mg = m\frac{v^2}{r}$

where

$\mu$ is the coefficient of static friction between the tires and the road

m is the mass of the car

g is the gravitational acceleration

v is the speed of the car

r is the radius of the curve

Re-arranging the equation,

$v=\sqrt{\mu gr}$

And by substituting the data of the problem, we find the speed at which the car begins to skid:

$v=\sqrt{(0.350)(9.8 m/s^2)(100 m)}=18.5 m/s$

7 0

3 years ago

Read 2 more answers

A negative ion of charge -2e is located at the origin and a second negative ion of charge -3e is located nearby at x = 3.8 nm ,

Rus_ich [418]

Answer:

$\vec{F}_{21}=-5.63\times 10^{-11}N\\\\\vec{F}_{21}=\\$

Explanation:

Given that

$Q_1 = -2e\, C\\\\Q_2=-3e\,C\\\\x= 3.8 \times 10^{-9}\,m\\\\y= 3.2 \times 10^{-9}\,m\\\\r=\sqrt{x^2+y^2}\\\\r= 4.96\times 10^{-9} m\\$

As both charges are negative so there exist force of repulsion in direction as shown in figure.

$F_{12}=\frac{kQ_1Q_2}{r^2}\\\\F_{12}= \frac{(9\times 10^9)(6)(1.602\times 10^{-19})^2}{(4.96\times 10^{-9})^2}\\\\F_{12}=5.63\times 10^{-11}N$

Angle at which force F12 is acting is

$\theta=tan^{-1}\frac{3.2}{3.8}\\\\\theta=tan^{-1}\frac{y}{x}\\\\\theta= 40.1^o$

$F_{x}=F_{12}cos\theta\\\\F_{x}=(5.63\times 10^{-11})cos(40.1)\\\\F_{x}=4.306\times 10^{-11}N\\\\F_{y}=F_{12}sin\theta\\\\F_{y}=(5.63\times 10^{-11})sin(40.1)\\\\F_{y}=3.62\times 10^{-11}N\\\\$

$\vec{F}_{12}=\vec{F}_{x}+\vec{F}_y\\\\\vec{F}_{12}=4.30\times 10^{-11}\,\hat{i} + 3.62\times 10^{-11}\,\hat{j}\\\\\vec{F}_{12}=$

Force exerted on charge -2e is equal in magnitude to F12 but is in opposite direction

$F_{21}=-5.63\times 10^{-11}N$

$\vec{F}_{21}=$

7 0

3 years ago

An imaginary line perpendicular to a reflecting surface is called ____refrac_____.

umka2103 [35]

Not totally sure but i would say a normal? its not refraction or incidence if its perpendicular and i dont think its a mirror if its an imaginary line so yeah normal (normals are always perpendicular to their surface too i think so)

4 0

3 years ago

A bus travelling 30.0 km/h east has a constant increase in speed of 1.5 m/s. What is its velocity 6.8s later

alekssr [168]

Answer:

8.72 seconds before

Explanation:

and 27.1 secs after

7 0

3 years ago