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svetlana [45]
3 years ago
5

What force is needed to accelerate an object 5 m/s if the object has a mass of 10kg?

Physics
1 answer:
nevsk [136]3 years ago
6 0
The <span>force that is needed to accelerate an object 5 m/s if the object has a mass of 10kg 50N because you multiply 5 and 10</span>
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Hi, I am having issues with this physics question. I find it quite complicated to solve and I don't even have the answer to this
galina1969 [7]

Answer:

200 m

Explanation:

Both Jerry and Tom start at the end of the pipe where the cheese is.  Jerry then runs into the pipe at 20 m/s while Tom chases him at 10 m/s.  When Jerry reaches the other end of the pipe, where Tom's cousin is, he turns around.  Eventually, he reaches Tom again, so he turns around again.  This continues until both Tom and Jerry reach the end of the pipe where Tom's cousin is.

We need to find the distance traveled by Jerry.  We already know Jerry's speed, so we just need to find the time.  We can do that using Tom's speed and the length of the pipe.

x = vt

100 m = (10 m/s) t

t = 10 s

So both Tom and Jerry run for 10 seconds.  Since Jerry runs at a speed of 20 m/s, the distance he travels is:

x = vt

x = (20 m/s) (10 s)

x = 200 m

Jerry runs a distance of 200 meters.

8 0
3 years ago
A playground is on the flat roof of a city school, 6.2m above the street below. The vertical wall of the building is h=7.30m hig
olya-2409 [2.1K]

Answer:

a. 18.13m/s

b. 0.84m

c. 2.4m

Explanation:

a. to find the speed at which the ball was lunched, we use the horizontal component.Since the point distance from the base of the ball is 24m and it takes 2.20 secs to reach the wall,we can say that

t=distance /speed

speed v_{x}=vcos\alpha=24/2.2=10.9m/s\\V=10.9/cos53^{0}\\V=18.13m/s

Hence the speed at which ball was lunched is 18.13m/s

b. from the equation

y=v_{y}t-\frac{1}{2}gt^{2}\\ v_{y}=18.13sin53=14.48m/s\\\\y=(14.48*2.2)-\frac{1}{2}9.8*2.2^{2}\\y=8.14m\\

the vertical distance at which the ball clears the wall is

y=8.14-7.3=0.84m

c. the time it takes the ball to reach the 6.2m vertically

6.2=14.47t-4.8t^{2}\\\\14.47t-4.8t^{2}-6.2=0\\\using\\t=-b±\frac{\sqrt{b^{2}-4ac} }{2a}\\ where \\a=-4.8, b=14.47t, c=6.2\\hence t=2.4secs

the horizontal distance covered at this speed is

y+24=(18.13cos53)2.4\\y=26.186-24\\y=2.12m

4 0
3 years ago
In which decade was Sputnik I launched? A. 1930-1939 B. 1950-1959 C. 1960-1969 D. 1980-1989
Olenka [21]
B) 1950-1959
Sputnik I was launched on October 4, 1957
6 0
3 years ago
5
Tanya [424]

Answer:

Reset

Explanation:

Digital methods are the methods that are uses methodological outlook to study societal change and cultural condition of online data. Reset is use to disguise data In digital methods. It is use to set again and conceal data by giving the data a different form. It restores the device to the original manufacture's settings.

8 0
3 years ago
Consider a stone in free fall on a planet with gravitational acceleration 3.4 m/s^2. Suppose you would like the stone to experie
Stella [2.4K]

Answer:

Angle of incline is 20.2978°

Explanation:

Given that;

Gravitational acceleration on a planet a = 3.4 m/s²

Gravitational acceleration on Earth g = 9.8 m/s²

Angle of incline = ∅

Mass of the stone = m

Force on the stone along the incline will be;

F = mgSin∅

F = ma

The stone has the same acceleration as that of the gravitational acceleration on the planet.

so

ma = mgSin∅

a = gSin∅

Sin∅ = a / g

we substitute

Sin∅ = (3.4 m/s²) / (9.8 m/s²)

Sin∅ = 0.3469

∅ = Sin⁻¹( 0.3469 )

∅ = 20.2978°

Therefore, Angle of incline is 20.2978°

8 0
3 years ago
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