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2 years ago

98 Points and brainlyest for 5 Science questions please I need it doe before 2:30 ET!!!

Physics

1 answer:

Marina CMI [18]2 years ago

5 0

Picture #1:
GPE = (mass) x (gravity) x (height)
GPE = (2 kg) x (9.8 m/s²) x (40 m) = 784 joules

KE = (1/2) (mass) (speed²)
KE = (1/2) (2 kg) (5 m/s)²
KE = (1 kg) (25 m²/s²) = 25 joules

Picture #2:
KE = (1/2) (mass) (speed²)
KE = (1/2) (2 kg) (10 m/s)²
KE = (1 kg) (100 m²/s²) = 100 joules

Picture #3:
GPE = (mass) x (gravity) x (height)
GPE = (20 kg) x (9.8 m/s²) x (2 m) = 392 joules

KE = (1/2) (mass) (speed²)
KE = (1/2) (20 kg) (5 m/s)²
KE = (10 kg) (25 m²/s²) = 250 joules

Picture #4:
GPE = (mass) x (gravity) x (height)
98 joules = (1 kg) x (9.8 m/s²) x (height)
Height = (98 joules) / (1 kg x 9.8 m/s²)
Height = 10 meters

Picture #5:
GPE = (mass) x (gravity) x (height)
39,200 Joules = (mass) x (9.8 m/s²) x (20 m)
Mass = (39,200 joules) / (9.8 m/s² x 20 m)
Mass = 200 kg

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A body is thrown vertically upward. Its velocity keep on decreasing. What happens to its kinetic energy when it reaches the maxi

Dmitrij [34]

Answer:

Explanation:

When a body is thrown upwards, its velocity decreases. This is because the kinetic energy gradually changes into potential energy. At the highest point, the velocity becomes zero since the kinetic energy gets completely converted into potential energy.

Hope this helps

plz mark as brainliest!!!!!!

7 0

2 years ago

A 70 kg hunter, standing on frictionless ice, shoots a 42 g bullet horizontally at a speed of 590 m/s . Part A Part complete Wha

Kisachek [45]

Answer:

The recoil velocity is 0.354 m/s.

Explanation:

Given that,

Mass of hunter = 70 kg

Mass of bullet = 42 g = 0.042 kg

Speed of bullet = 590 m/s

We need to calculate the recoil speed of hunter

Using conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

Where, $m_{1}$ = mass of hunter

$m_{2}$ = mass of bullet

u = initial velocity

v = recoil velocity

Put the value in the equation

$0+0=70\times v_{1}+0.042\times590$

$v_{1}=-\dfrac{0.042\times590}{70}$

$v=-0.354\ m/s$

Hence, The recoil velocity is 0.354 m/s.

8 0

2 years ago

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kkurt [141]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

Estimate the volume of a typical house (2050 feet squared in size and 10 feet tall) answer in units of meters squared

Aloiza [94]

Answer:

$Volume = 6248.48 m^{3}$

Explanation:

Given:

The area of the house $A = 2050\ ft^{2}$

The height of the house $h=10\ ft$

We need to find the volume of a typical house.

Solution:

We find the volume of the house by multiplying the area of the house and height of the house.

$Volume = Area\times height$

$Volume = A\times h$

Area and height of the house are known, so we substitute these value in above equation.

$Volume = 2050\times 10$

$Volume = 20500\ ft^{3}$

Now we convert the unit from feet to meter.

Divide the volume by 3.2808 for $m^{3}$

$Volume = \frac{20500}{3.2808}$

$Volume = 6248.48\ m^{3}$

Therefore, the volume of the house is $6248.48 m^{3}$

8 0

2 years ago

A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on

Olin [163]

The crate moves at constant velocity, this means that its acceleration is zero, so the net force acting on the crate is zero (Newton's second law).

There are only two forces acting on the crate: the force F applied by the worker and the frictional force, acting in the opposite direction: $\mu m g$ , where $\mu=0.25$ is the coefficient of friction and $m=30.0 kg$ is the mass of the crate. Since the net force should be equal to zero, the two forces must have same magnitude, so we have:
$F=\mu m g=(0.25)(30.0 kg)(9.81 m/s^2)=73.8 N$
And so, this is the force that the worker must apply to the crate.

5 0

2 years ago