Ask question

Ask question

All categories

3 years ago

13

A tiger leaps with an initial velocity of 35.0 km/hr at an angle of 13.0ᶿ with respect to the horizontal. What are the component

s of the tiger’s velocity?

1 answer:

Lorico [155]3 years ago

3 0

Answer:

Vx = 35 x cos(13deg)

Vy = 35 x sin(13deg) - gt

(g is acceleration due to gravity =~9.8 meter/second^2, t is time in second)

Explanation:

The tiger leaps up, then x and y component of its velocity are:

Vx = Vo x cos(alpha)

Vy = Vo x sin(alpha) - gt

(Vo is tiger's initial velocity, alpha is angle between its leaping direction and horizontal plane)

Hope this helps!

You might be interested in

2 Jupiter orbits the sun in a nearly circular path with radius 7.8x10^11m. The orbital period of Jupiter is 12 years.

bazaltina [42]

Mass of Jupiter=1.9×10

27

㎏=M

1

Mass of Sun=1.99×10

30

㎏=M

2

Mean distance of Jupiter from Sun=7.8×10

11

m=r

G=6.67×10

−11

N㎡㎏

−2

Gravitational Force, F=

r

2

GM

1

M

2

F=

(7.8×10

11

)

2

6.67×10

−11

×1.9×10

27

×1.99×10

30

F=4.16×10

23

N

4 0

2 years ago

A car starts from rest and accelerates uniformly over a time of 7.25 seconds for a distance of 210 m. Determine the acceleration

lara31 [8.8K]

Answer:

Acceleration is 7.990487515m/s²

Initial velocity is 0m.s

Explanation:

s=ut+(1/2)at²

210=0(7.25)+(1/2)a(7.25²)

210=26.28125a

∴a=7.990487515m/s²

'Vi' or 'u' is the inital speed. Since it starts from rest, this equals 0.

8 0

2 years ago

To win a prize at the county fair, you're trying to knock down a heavy bowling pin by hitting it with a thrown object. Should yo

Setler [38]

Answer:

Being an elastic object, rubber ball will be an ideal choice as it will bounce off the bowling pit and will experience a large change in momentum in comparison with the beanbag which will either slow down or come to a halt upon hitting a bowling pit. That is why rubber ball will experience a greater impulse and the bowling pin will experience the negative impulse of the rubber ball.

For Rubber Ball

Upon elastic collision it will reverses the direction and move with velocity equal or less then original

change in momentum = P

$P = m(v_{f} -v_{i})\\v_{f}=-v_{i} \\ P = -2mv_{i}$

For Beanbag

value of impulse will large if velocity is zero.

$v_{f}=0\\ P = -mv_{i}$

Explanation:

8 0

2 years ago

Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o

Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 6.05x10^{-14}m$ < $\lambda_{electron} = 1.10x10^{-10}m$

Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

Part A

Case for the electron:

$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

$\lambda = 6.05x10^{-14}m$

Hence, the proton has a smaller wavelength than the electron.

<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Case for the electron:

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

$v = 1.316x10^{8}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

$\lambda = 5.525x10^{-12}m$

Case for the proton :

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}$

$v = 3.07x10^{6}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

Hence, the proton has a smaller wavelength than the electron.

7 0

2 years ago

A driver's foot presses with a steady force of 20N on a pedal in a car as shown.

azamat

Answer:

160N

Explanation:

Moments must be conserved - so.

$20 * 0.4 = F * 0.05$

$F=\frac{20*0.4}{0.05} = 160$

6 0

3 years ago