All categories

3 years ago

When a projectile reaches the highest point the vertical component of the acceleration is:

Physics

1 answer:

Mazyrski [523]3 years ago

6 0

Answer:

The acceleration is g.

Taking the upward direction as positive

V = Vy y - 1/2 g t^2

Taking the downward direction as positive

V = -V y + 1/2 g t^2

One can choose either direction as positive, but the acceleration is

the same as g (it is g) while the projectile is in the air.

You might be interested in

Physical values in the real world have two componnts: magnitude and ____

Dahasolnce [82]

Answer:

DIRECTION

Explanation:

3 0

3 years ago

A force of 150 N is applied to a 25 cm- piston in a hydraulic machine. What is the area

maxonik [38]

Answer:

8.3 cm^2

.................

6 0

3 years ago

A rocket is fired at 100 m/s at an angle of 37, what was its speed at the top of its path?

Nadusha1986 [10]

Answer:

0m/s

Explanation:

Since its fired at an angle, at the top there will be a split second where the velocity will be 0, as it has a parabolic shape, so the speed at the top of its path is 0

4 0

3 years ago

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

3 years ago

What is its rotational inertia about the meter stick in the

Sindrei [870]

Answer:

Honestly I cant understand that so please explain it someone

Explanation:

3 0

3 years ago