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jasenka [17]
3 years ago
5

What are ways to improve the design of this experiment? Check all that apply

Physics
1 answer:
Serhud [2]3 years ago
3 0

Answer:

B,D,E

Explanation:

I got you

B. Experiment with a wider range of materials.

D.Use a laboratory galvanometer to make precise measurements.

E. Test the strength of the electromagnet by varying the number of wire coils.

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An object weighs 32 newtons. What is its mass if a gravitometer indicates that g = 8.25 m/s?
Elan Coil [88]

Explanation:

weight=mass×g

32=mass×8.25

mass=

\frac{128}{33}  = 3.878kg

8 0
2 years ago
A 1200-kilogram automobile in motion strikes a 0.0001-kilogram insect. As a result, the insect is accelerated at a rate of 100 m
nadya68 [22]

Answer:

force = 1 × 10^{-2} N

Explanation:

given data

automobile mass = 1200 kg

insect mass = 0.0001 kg

insect accelerated = 100 m/s²

to find out

magnitude of the force the insect exerts on the car

solution

we get here force the insect exerts that is express as

force = mass × acceleration    ............1

put here value we get

force = 0.0001 × 100 m/s²

force = 1 × 10^{-2} N

6 0
3 years ago
Read 2 more answers
if a man has a mass of 83 kilograms on earth, what will the force of gravity on his body be on the moon
AnnZ [28]

If a man has a mass of 83 kilograms on Earth, the force of gravity on his body be on the moon 135.6N. force =mass*acc , 83 * 9.8/6= 813.4/6 = 135.6N

3 0
3 years ago
Read 2 more answers
A projectile is launched into the air with the initial speed of vi = 40 m/s at a launch angle of 20 degrees above the horizontal
Sphinxa [80]

The range of the projectile is 188 m

Explanation:

The motion of the arrow in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction  

The path of a projectile is the combination of these two motions: see figure in attachment.

In order to find the horizontal range of the projectile, we just need to calculate the horizontal distance travelled.

We have:

t = 5.0 s (time of fligth of the projectile)

and the horizontal velocity is constant, and it is given by

v_x = v_i cos \theta

where

v_i = 40 m/s is the initial velocity

\theta=20^{\circ} is the angle of projection

Substituting,

v_x = (40)(cos 20^{\circ})=37.6 m/s

And therefore, the range of the projectile is:

d=v_x t = (37.6)(5.0)=188 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

5 0
3 years ago
Block A has a mass of 0.5kg, and block B has a mass of 2kg. Block is is released at a height of 0.75 meters above B. The coeffic
VikaD [51]

Answer:

0.075 m

Explanation:

The picture of the problem is missing: find it in attachment.

At first, block A is released at a distance of

h = 0.75 m

above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:

m_Agh=\frac{1}{2}m_Av_A^2

where

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

m_A=0.5 kg is the mass of the block

v_A is the speed of the block A just before touching block B

Solving for the speed,

v_A=\sqrt{2gh}=\sqrt{2(9.8)(0.75)}=3.83 m/s

Then, block A collides with block B. The coefficient of restitution in the collision is given by:

e=\frac{v'_B-v'_A}{v_A-v_B}

where:

e = 0.7 is the coefficient of restitution in this case

v_B' is the final velocity of block B

v_A' is the final velocity of block A

v_A=3.83 m/s

v_B=0 is the initial velocity of block B

Solving,

v_B'-v_A'=e(v_A-v_B)=0.7(3.83)=2.68 m/s

Re-arranging it,

v_A'=v_B'-2.68 (1)

Also, the total momentum must be conserved, so we can write:

m_A v_A + m_B v_B = m_A v'_A + m_B v'_B

where

m_B=2 kg

And substituting (1) and all the other values,

m_A v_A = m_A (v_B'-2.68) + m_B v_B'\\v_B' = \frac{m_A v_A +2.68 m_A}{m_A + m_B}=1.30 m/s

This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to

\frac{1}{2}m_B v_B'^2 = \frac{1}{2}kx^2

where

k = 600 N/m is the spring constant

x is the compression of the spring

And solving for x,

x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(2)(1.30)^2}{600}}=0.075 m

5 0
3 years ago
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