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german
3 years ago
8

NEED HELP NOW WILL GIVE BRAINLIEST

Physics
2 answers:
umka21 [38]3 years ago
8 0
It’s a, b, and d..................................
Arturiano [62]3 years ago
6 0

Answer:

B. solar flares sunspots,  solar flares prominences

Explanation:

Some features of the Sun's surface include sunspots, solar flares, and prominences

pls mark me brainliest its the right answer

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Suppose that a block is pulled 16 meters across a floor. What amount of work is done if the force used to drag the block is 22 N
klemol [59]
We have: A= F.s
So, A = 22 x 16
A= 352J
Só A is correct answer
3 0
3 years ago
Read 2 more answers
Technically, how would you know if any work was done on an object?
kherson [118]

Answer:

through reading the question carefully and using yourself as an exampke

8 0
3 years ago
A 1100kg car pulls a boat on a trailer. (a) what total force resists the motion of the car, boat,and trailer, if the car exerts
likoan [24]
Refer to the figure shown below.

m₁ = 1100 kg, the mass of the car
m₂ = 700 kg, the mass of the trailer and boat
F = 1900 N, the driving force acting on the car
N₁ = m₁g,  the normal reaction on the car
N₂ = m₂g, the normal force on the trailer and boat
μN₁ and μN₂ are frictional forces, where  =  kinetic coefficient of friction
T = the force in the hitch between the car and trailer.

Part (a)
Let R₁ = the total force that resists the motion of the car, boat, and trailer.
Because the acceleration is 0.550m/s², therefore
(m₁ + m₂ kg)*(0.55 m/s²) = F
(1100+700 kg)*(0.55 m/s²) = (1900 - R₁) N
990 = 1900 - R
R₁ = 910 N

Answer: The resistive force is 910 N

Part (b)
80% of the resistive forces are experienced by the boat and trailer.
Let the resistive force be R₂.
Then
R₂ = 0.8*R₁ = 728 N
If the tension in the hitch between the car and the trailer is T, then
T - R₂ = m₂(0.55 m/s²)
(T - 728 N) = (700 kg)*(0.55 m/s²)
T - 728 = 385
T = 1113 N

Answer: The force in the hitch is 1113 N

3 0
3 years ago
You are standing in a building whose height (40m) you throw a ball downward at a angle of -30 at a speed of (10m/s) acceleration
jeyben [28]

Answer: 3.41 s

Explanation:

Assuming the question is to find the time t the ball is in air, we can use the following equation:

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

Where:

y=0m is the final height of the ball

y_{o}=40 m is the initial height of the ball

V_{o}=10 m/s is the initial velocity of the ball

t is the time the ball is in air

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

\theta=30\°

Then:

0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}

0 m=40 m+5m/s t-4.9 m/s^{2}t^{2}

Multiplying both sides of the equation by -1 and rearranging:

4.9 m/s^{2}t^{2}-5m/s t-40 m=0

At this point we have a quadratic equation 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}

Where:

a=4.9

b=-5

c=-40

Substituting the known values:

t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}

Solving the equation and choosing the positive result we have:

t=3.41 s  This is the time the ball is in air

5 0
4 years ago
A 115 kg hockey player, Adam, is skating east whenever he tackles a 133 kg player, Bob, skating west at -1.59 m/s. Afterwards, t
snow_tiger [21]

Answer:

2.38 m/s

Explanation:

Momentum is conserved:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

After the collision, they move at the same speed, so v₁ = v₂ = v.

m₁u₁ + m₂u₂ = (m₁ + m₂) v

Plugging in numbers:

(115 kg) u₁ + (133 kg) (-1.59 m/s) = (115 kg + 133 kg) (0.250 m/s)

u₁ = 2.38 m/s

6 0
3 years ago
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