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2 years ago

For number 1 I'm really confused

Physics

2 answers:

Airida [17]2 years ago

7 0

According to Newton's second law of motion,

F = mass x acceleration = 2 x 3 = 6 N.

Hence 6 N of force is required.

Hope this helps!

marishachu [46]2 years ago

4 0

The answer is 6{N}. :)

You might be interested in

Is is false or true The closer objects are, the stronger the gravitational force between them.

____ [38]

Answer:

TRUE

Explanation:

The gravitational force between two objects becomes weaker if the two objects are moved apart and stronger if they are brought closer together; that is, the force depends on the distance between the objects

5 0

3 years ago

a linear function has the same y-intercept as x + 4y equals 16 and it's graph contains the point (4,5). Find the slope of the li

navik [9.2K]

Answer: $\bold{\text{Slope (m)}=\dfrac{1}{4}}$

Explanation:

A linear equation is of the form: y = mx + b where

m is the slope
b is the y-intercept (where it crosses the y-axis)

x + 4y = 16

4y = -x + 16

$y = -\dfrac{1}{4}x+\dfrac{16}{4}$

$y=-\dfrac{1}{4}x+4$

The y-intercept (b) = 4

Next, find the slope given point (4, 5) and b = 4

$y=mx+b\\\\5=m(4)+4\\\\1=4m\\\\\dfrac{1}{4}=m\\\\\\\\\large\boxed{Slope (m)=\dfrac{1}{4}}$

6 0

3 years ago

If a person is walking at 1.2 m/s and 60 seconds later the person is running at 10 m/s, what was the acceleration rate?

Marina CMI [18]

The acceleration rate would be .14667 m/s^2

6 0

2 years ago

Suppose that on earth you can throw a ball vertically upward a distance of 1.20 m. Given that the acceleration of gravity on the

tatuchka [14]

Answer:

7.04 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0

s = Displacement on Earth = 1.2 m

a = Acceleration due to gravity on Moon = 1.67 m/s²

a = Acceleration due to gravity Earth= 9.81 m/s²

Accelration going up is considered as negetive

Initial Velocity of the ball

$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 1.2-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 1.2}\\\Rightarrow u=4.85\ m/s$

Assuming that the ball is thrown with the same velocity on the Moon, displacement of the ball is

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-4.85^2}{2\times -1.67}\\\Rightarrow s=7.04\ m$

The displacement of the ball on the moon is 7.04 m

6 0

3 years ago

Please help me with this (with explanation)

Sergeeva-Olga [200]

Suppose the cyclist travels for a total time of t hours.

For 20 min = 1/3 hr, the cyclist does not move.

Over the remaining (t - 1/3) hr, the cyclist is moving at a constant speed of 22.0 km/hr, so that the cyclist would travel a distance of

x = (22.0 km/hr) • ((t - 1/3) hr) ≈ (22.0 km/hr) t - 7.33 km

If the cyclist's average speed over the total time t was 17.5 km/hr, then by the definition of average speed,

17.5 km/hr = x / t

Replace x with the distance expression from earlier:

17.5 km/hr = ((22.0 km/hr) t - 7.33 km) / t

Solve for t :

17.5 km/hr = 22.0 km/hr - (7.33 km) / t

(7.33 km) / t = 4.5 km/hr

t = (7.33 km) / (4.5 km/hr)

t ≈ 1.62963 hr

Then the distance the cyclist traveled must have been

x ≈ (22.0 km/hr) (1.62963 hr) - 7.33 km ≈ 28.5 km

and so the answer is A.

Alternatively, as soon as you arrive at

17.5 km/hr = x / t

you can instead solve for t in terms of x, then plug that into the distance equation.

t = x / (17.5 km/hr)

then

x ≈ (22.0 km/hr) (x / (17.5 km/hr)) - 7.33 km

x ≈ 1.25714 x - 7.33 km

0.25714x ≈ 7.33 km

x = (7.33 km) / 0.25714 ≈ 28.5 km

6 0

2 years ago