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ser-zykov [4K]
3 years ago
6

Three forces of magnitude 10N, 5N and 4N act on an object in the directions North, West and East respectively. Find the magnitud

e and directions of their resultant
Physics
1 answer:
Gekata [30.6K]3 years ago
8 0

Answer:

19N to the south

Explanation:

F =10N + 5N + 4N

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A person pushes on a box with a force of 500N. If the box does not move, what is the force of static friction on the box?
Alexxx [7]

Answer:

c

Explanation:

4 0
3 years ago
The half-life of the radioactive element beryllium-13 is 5 × 10-10 seconds, and half-life of the radioactive element beryllium-1
telo118 [61]
<h2>Answer: The half-life of beryllium-15 is 400 times greater than the half-life of beryllium-13.</h2>

Explanation:

The half-life h of a radioactive isotope refers to its decay period, which is the average lifetime of an atom before it disintegrates.

In this case, we are given the half life of two elements:

beryllium-13: h_{B-13}=5(10)^{-10}s=0.0000000005s

beryllium-15: h_{B-15}=2(10)^{-7}s=0.0000002s

As we can see, the half-life of beryllium-15 is greater than the half-life of beryllium-13, but how great?

We can find it out by the following expression:

h_{B-15}=X.h_{B-13}

Where X is the amount we want to find:

X=\frac{h_{B-15}}{h_{B-13}}

X=\frac{2(10)^{-7}s}{5(10)^{-10}s}

Finally:

X=400

Therefore:

The half-life of beryllium-15 is <u>400 times greater than</u> the half-life of beryllium-13.

8 0
3 years ago
A child on a skateboard experiences a 75 N force with an acceleration of 1.5 m/s2.
cupoosta [38]

Answer:

The mass of the child + skateboard is 50 kg

Explanation:

In this problem, we can apply Newton's second law:

F = ma

where

F is the net force on a system

m is the mass of the system

a is the acceleration of the system

In this problem, our system is the child + the skateboard. The net force on them is

F = 75 N

and their acceleration is

a=1.5 m/s^2

So we can re-arrange the equation above to find their combined mass:

m=\frac{F}{a}=\frac{75}{1.5}=50 kg

3 0
3 years ago
An object thrown vertically upward from the surface of a celestial body at a velocity of 36 ​m/s reaches a height of sequalsminu
Anarel [89]

Answer:

v = -1.8t+36

20 seconds

360 m

40 seconds

36 m/s

The object speed will increase when it is coming down from its highest height.

Explanation:

s=-0.9t^2+36t

Differentiating with respect to time we get

\frac{ds}{dt}=-1.8t+36\\\Rightarrow v=-1.8t+36

a) Velocity of the object after t seconds is v = -1.8t+36

At the highest point v will be 0

0=-1.8t+36\\\Rightarrow t=\frac{-36}{-1.8}\\\Rightarrow t=20\ s

b) The object will reach the highest point after 20 seconds

s=-0.9t^2+36t\\\Rightarrow s=-0.9\times 20^2+36\times 20\\\Rightarrow s=360\ m

c) Highest point the object will reach is 360 m

s=-0.9t^2+36t\\\Rightarrow 360=-0.9t^2+36t\\\Rightarrow -0.9t^2+36t-360=0\\\Rightarrow -9t^2+360t-3600=0

\frac{-360\pm \sqrt{0}}{2\left(-9\right)}\\\Rightarrow t=20\ s

d) Time taken to strike the ground would be 20+20 = 40 seconds

[tex]v=u+at\\\Rightarrow v=0+0.9\times 2\times 20\\\Rightarrow v=36\ m/s

Acceleration will be taken as positive because the object is going down. Hence, the sign changes. 2 is multiplied because the expression is given in the form of s=ut+\frac{1}{2}at^2

e) The velocity with which the object strikes the ground will be 36 m/s

f) The speed will increase when the object has gone up and for 20 seconds and falls down for 20 seconds. The object speed will increase when it is coming down from its highest height.

4 0
3 years ago
Select the statement that is not true about the Hubble Telescope:
xxMikexx [17]

Answer:

The option is B is not true for Hubble telescope.

4 0
3 years ago
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