All categories

3 years ago

Can Hockey players hold the puck with their open hand?

Physics

1 answer:

devlian [24]3 years ago

8 0

Answer:

Yes they can stop or bat the puck with their open hand

Explanation:

You might be interested in

A car is traveling in a straight path due east with a velocity of 8 meters per second. In a 10 second interval, how many meters

34kurt

The answer is c 18meters because if u add you get that answer hope I'm right bye

6 0

3 years ago

Read 2 more answers

5)List an appropriate SI base unit (with a prefix as needed ) for the following:

AlexFokin [52]

B) kg
d) cm
e)kg
g) km
h) kg

5 0

3 years ago

Read 2 more answers

(a) What resonant frequency would you expect from blowing across the top of an empty soda bottle that is 18 cm deep, if you assu

V125BC [204]

Answer:

476.387 Hz

714.583 Hz

Explanation:

L = Length of tube

v = Speed of sound in air = 343 m/s

Frequency for a closed tube is given by

$f=\dfrac{v}{4L}\\\Rightarrow f=\dfrac{343}{4\times 0.18}\\\Rightarrow f=476.389\ Hz$

The frequency is 476.387 Hz

If it was one third full $L=0.18-\dfrac{1}{3}0.18$

$f=\dfrac{v}{4L}\\\Rightarrow f=\dfrac{343}{4\times (0.18-\dfrac{1}{3}0.18)}\\\Rightarrow f=714.583\ Hz$

The frequency is 714.583 Hz

5 0

3 years ago

B-1:

zalisa [80]

Answer:

(a) t = 1.67 s

(b) s₂ = 45 m

Explanation:

Here, we use the formula:

s = vt

FOR Seth:

s₁ = v₁t₁

where,

s₁ = distance covered by Seth

v₁ = speed of Seth = 9 m/s

t₁ = time taken by Seth

FOR Mack:

s₂ = v₂t₂

where,

s₂ = distance covered by Mack

v₂ = speed of Mack = 27 m/s

t₂ = time taken by Mack

since, initially Mack is 30 m behind Seth. Therefore,

(a)

s₂ = s₁ + 30 m

using formulae:

v₂t₂ = v₁t₁ + 30 m

but, the time of catching is same for both (t₁ = t₂ = t)

v₂t = v₁t + 30 m

using values:

(27 m/s)t - (9 m/s)t = 30 m

t = (30 m)/(18 m/s)

(b)

s₂ = v₂t

using values:

s₂ = (27 m/s)(1.67 s)

3 0

3 years ago

To test a slide at an amusement park, a block of wood with mass 3.00 kgkg is released at the top of the slide and slides down to

dem82 [27]

Answer:

511.1 J

Explanation:

We are given that

Mass of wood block=m=3 kg

Vertical distance,h=23 m

Horizontal distance =x=30 m

Distance traveled in downward direction y=40 m

Initial velocity,u=0

$y=ut+\frac{1}{2} gt^2$

Where $g=9.8 m/s^2$

$40=0+\frac{1}{2}(9.8)t^2=4.9t^2$

$t^2=\frac{40}{4.9}$

$t=\sqrt{\frac{40}{4.9}}=2.86 s$

$x=v_x\times t$

$30=v_x(2.86)$

$v=v_x=\frac{30}{2.86}=10.49 m/s$

By work energy theorem

Change in kinetic energy=Work done= mgh-W

$\frac{1}{2}mv^2-\frac{1}{2}mu^2=3\times 9.8\times 23-W$

$\frac{1}{2}(3)(10.49)^2-0=676.2-W$

$W=676.2-\frac{1}{2}(3)(10.49)^2=511.1 J$

Hence, the work done due to friction on the block as it slides down the ramp=511.1 J

6 0

3 years ago