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prisoha [69]
3 years ago
12

(1) Arectangularmetal block with mass 18 kg has

Physics
1 answer:
sleet_krkn [62]3 years ago
8 0

Answer:

Volume=0.003m^3 density=6000kg/m^3

Explanation:

mass=18kg

volume=length x width x height

Volume of block=0.1x0.15x0.2

volume of block=0.003m^3

Density=mass ➗ volume

Density=18 ➗ 0.003

Density=6000kg/m^3

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a ball is projected upward at time t = 0.00 s from a point on a roof 70 m above the ground. The ball rises, then falls and strik
grin007 [14]

Answer: 17.68 s

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:  

y=y_{o}+V_{o}t-\frac{1}{2}gt^{2} (1)  

Where:  

y=0 is the height of the ball when it hits the ground  

y_{o}=70 m is the initial height of the ball

V_{o}=82m/s is the initial velocity of the ball  

t is the time when the ball strikes the ground

g=9.8m/s^{2} is the acceleration due to gravity  

Having this clear, let's find t from (1):  

0=70m+(82m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2} (2)  

Rewritting (2):

-\frac{1}{2}(9.8m/s^{2})t^{2}+(82m/s)t+70m=0 (3)  

This is a quadratic equation (also called equation of the second degree) of the form at^{2}+bt+c=0, which can be solved with the following formula:

t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}  (4)

Where:

a=-\frac{1}{2}(9.8m/s^{2}

b=82m/s

c=70m

Substituting the known values:

t=\frac{-82 \pm \sqrt{82^{2}-4(-\frac{1}{2}(9.8)(70)}}{2a}  (5)

Solving (5) we find the positive result is:

t=17.68 s

7 0
3 years ago
In a simple electric circuit, a 110-volt electric heater
Readme [11.4K]

Answer: 55 ohms

Explanation:

Given that,

Voltage of heater (v) = 110-volt

Current drawn by heater (I) = 2.0 amperes

resistance of the heater (r) = ?

Since voltage, current and resistance are involved, apply the formula for ohms law.

Voltage = current x resistance

i.e v = ir

where r = v / i

r = 110 volts / 2.0 A

r = 55 ohms

Thus, the resistance of the heater is 55 ohms

6 0
3 years ago
What is the impulse of a dodgeball hitting a person in the face if it's initial velocity was 10m/s and it's velocity after hitti
Alborosie

Answer:)

Explanation:)

6 0
3 years ago
Which bright solar feature is shown in the picture above?
Ghella [55]

Answer : (B) Prominence

Explanation :

A large, glittering and gaseous characteristic which is extending outward from the surface of the sun is called <em>Prominence</em>.

<em>Photosphere</em> is one of the layer of sun where the prominence are anchored and then they move into the corona of the sun.

<em>Corona</em> is a region in the surface of the sun which is the constituent of hot ionized gases (plasma).

The prominence consists of colder plasma and this prominence plasma is much more shining and denser as compared to coronal plasma.

Hence, the correct option is (B) Prominence.

6 0
4 years ago
Read 2 more answers
A weightlifter lifts a 250-kg mass 0.5 meters above his head, how much PEg does the mass have (Note: g=9.8 m/s2)? Round your ans
morpeh [17]

Answer:

1225 J

Explanation:

The Gravitational potential energy (PEG) gained by a mass lifted above the ground is given by

PE=mgh

where

m is the mass

g = 9.8 m/s^2 is the acceleration due to gravity

h is the height at which the object has been lifted

In this problem, we have

m = 250 kg

h = 0.5 m

So, the PE of the object is

PE=(250 kg)(9.8 m/s^2)(0.5 m)=1225 J

6 0
3 years ago
Read 2 more answers
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