A cheetah runs with a positive acceleration. What does this tell you about the velocity time graph for the cheetah?

An artillery shell is fired with an initial speed of 300m/s at 55 deegry about the horizontal,it explodes on a mountain side 42s

photoshop1234 [79]

Answer:

x = 7226.94 m

y = 1677.4 m

Explanation:

This is a classic problem of a parabolic move. Let's analyze the given data.

The speed is 300 m/s but it's fired at 55°, so the x and y component of the speed would be:

Vx = 300 cos55 = 172.07 m/s

Vy = 300 sin55 = 245.75 m/s

Now, to get the x and y components of the shell, we need to apply the following expressions:

X = Xo + Vx*t (1)

Y = Yo + Vy*t - 1/2 gt² (2)

Xo and Yo is 0, and we already have the speed in x and y, so the components are:

X = 0 + (172.07)(42)

For the y component:

Y = 0 + (245.75)(42) - 1/2 (9.8)(42)²

Y = 10321.5 - 8643.6

Hope this helps

8 0

3 years ago

In xray machines, electrons are subjected to electric fields as great as 6.0 x 10^5 N/C. Find

katrin2010 [14]

Answer:

a = 1.055 x 10¹⁷ m/s²

Explanation:

First, we will find the force on electron:

$E = \frac{F}{q}\\\\F = Eq\\$

where,

F = Force = ?

E = Electric Field = 6 x 10⁵ N/C

q = charge on electron = 1.6 x 10⁻¹⁹ C

Therefore,

$F = (6\ x\ 10^5\ N/C)(1.6\ x\ 10^{-19}\ C)\\$

F = 9.6 x 10⁻¹⁴ N

Now, we will calculate the acceleration using Newton's Second Law:

$F = ma\\a = \frac{F}{m}\\$

where,

a = acceleration = ?

m = mass of electron = 9.1 x 10⁻³¹ kg

therefore,

$a = \frac{9.6\ x\ 10^{-14}\ N}{9.1\ x\ 10^{-31}\ kg}\\\\$

5 0

3 years ago

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is

amm1812

Answer: $1.066\times 10^7 m/s$

Explanation:

Given

Charge per unit area on each plate( $\sigma$ )= $2.2\times 10^{-7}$

Plate separation(y)=0.013 m

and velocity is given by

$v^2-u^2=2ay$

where a=acceleration is given by

$a=\frac{F}{m}=\frac{eE}{m}$

e=charge on electron

E=electric field

m=mass of electron

$E=\frac{\sigma }{\epsilon _0}$

$a=\frac{e\sigma }{m\epsilon _0}$

substituting values

$v=sqrt{\frac{2e\sigma y}{m\epsilon _0}}$

$v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 2.2\times 10^{-7}\times 0.013}{9.1\times 10^{-31}\times 8.85\times 10^{-12}}}$

$v=1.066\times 10^7 m/s$

4 0

3 years ago

Can you cry underwater

Katen [24]

No we can not cry under water unless you want to cry drowning. Crying requires breathing

3 0

3 years ago

Which of these actions will increase friction? Select three options. Scratching a surface to make it rougher polishing a surface

k0ka [10]

Answer:

scratching a surface to make it rougher

increasing the size of a flying object

adding extra weight to an object

Explanation:

5 0

3 years ago

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