Ask question

Ask question

All categories

2 years ago

13

If Emily throws the ball at an angle of 30∘ below the horizontal with a speed of 14m/s, how far from the base of the dorm should

Allison stand to catch the ball? Assume the vertical distance between where Emily releases the ball and Allison catches it is 4.0m.

1 answer:

liubo4ka [24]2 years ago

8 0

Emily throws the ball at 30 degree below the horizontal

so here the speed is 14 m/s and hence we will find its horizontal and vertical components

$v_x = 14 cos30 = 12.12 m/s$

$v_y = 14 sin30 = 7 m/s$

vertical distance between them

$\delta y = 4 m$

now we will use kinematics in order to find the time taken by the ball to reach at Allison

$\delat y = v_y *t + \frac{1}{2} at^2$

here acceleration is due to gravity

$a = 9.8 m/s^2$

now we will have

$4 = 7 * t + \frac{1}{2}*9.8 * t^2$

now solving above quadratic equation we have

$t = 0.44 s$

now in order to find the horizontal distance where ball will fall is given as

$d = v_x * t$

here it shows that horizontal motion is uniform motion and it is not accelerated so we can use distance = speed * time

$d = 12.12 * 0.44 = 5.33 m$

so the distance at which Allison is standing to catch the ball will be 5.33 m

You might be interested in

A thin, horizontal, 18-cm-diameter copper plate is charged to -3.8 nC. Assume that the electrons are uniformly distributed on th

son4ous [18]

Answer:

Part a)

$E = 8436.7 N/C$

Part b)

$E = 8436.7 N/C$

Explanation:

Part a)

Electric field due to large sheet is given as

$E = \frac{\sigma}{2\epsilon_0}$

$\sigma = \frac{Q}{A}$

$Q = -3.8 nC$

$A = \pi(0.09)^2$

$A = 0.025 m^2$

$\sigma = \frac{-3.8\times 10^{-9}}{0.025}$

$\sigma = -1.5 \times 10^{-7} C/m^2$

now the electric field is given as

$E = \frac{-1.5 \times 10^{-7}}{2(8.85 \times 10^{-12})}$

$E = 8436.7 N/C$

Part b)

Now since the electric field is required at same distance on other side

so the field will remain same on other side of the plate

$E = 8436.7 N/C$

5 0

2 years ago

a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres

shusha [124]

Answer:

Part a) When collision is perfectly inelastic

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b) When collision is perfectly elastic

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

$mv_m = (m + M)v$

so we have

$v = \frac{mv_m}{m + M}$

now we know that in order to complete the circle we will have

$v = \sqrt{5Rg}$

$\frac{mv_m}{m + M} = \sqrt{5Rg}$

now we have

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b)

Now we know that collision is perfectly elastic

so we will have

$v = \frac{2mv_m}{m + M}$

now we have

$\sqrt{5Rg} = \frac{2mv_m}{m + M}$

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

6 0

2 years ago

A 5.00μF parallel-plate capacitor is connected to a 12.0 V battery. After the capacitor is fully charged, the battery is disconn

EastWind [94]

(a) 12.0 V

In this problem, the capacitor is connected to the 12.0 V, until it is fully charged. Considering the capacity of the capacitor, $C=5.00 \mu F$ , the charged stored on the capacitor at the end of the process is

$Q=CV=(5.00 \mu F)(12.0 V)=60 \mu C$

When the battery is disconnected, the charge on the capacitor remains unchanged. But the capacitance, C, also remains unchanged, since it only depends on the properties of the capacitor (area and distance between the plates), which do not change. Therefore, given the relationship

$V=\frac{Q}{C}$

and since neither Q nor C change, the voltage V remains the same, 12.0 V.

(b) (i) 24.0 V

In this case, the plate separation is doubled. Let's remind the formula for the capacitance of a parallel-plate capacitor:

$C=\frac{\epsilon_0 \epsilon_r A}{d}$

where:

$\epsilon_0$ is the permittivity of free space

$\epsilon_r$ is the relative permittivity of the material inside the capacitor

A is the area of the plates

d is the separation between the plates

As we said, in this case the plate separation is doubled: d'=2d. This means that the capacitance is halved: $C'=\frac{C}{2}$ . The new voltage across the plate is given by

$V'=\frac{Q}{C'}$

and since Q (the charge) does not change (the capacitor is now isolated, so the charge cannot flow anywhere), the new voltage is

$V'=\frac{Q}{C'}=\frac{Q}{C/2}=2 \frac{Q}{C}=2V$

So, the new voltage is

$V'=2 (12.0 V)=24.0 V$

(c) (ii) 3.0 V

The area of each plate of the capacitor is given by:

$A=\pi r^2$

where r is the radius of the plate. In this case, the radius is doubled: r'=2r. Therefore, the new area will be

$A'=\pi (2r)^2 = 4 \pi r^2 = 4A$

While the separation between the plate was unchanged (d); so, the new capacitance will be

$C'=\frac{\epsilon_0 \epsilon_r A'}{d}=4\frac{\epsilon_0 \epsilon_r A}{d}=4C$

So, the capacitance has increased by a factor 4; therefore, the new voltage is

$V'=\frac{Q}{C'}=\frac{Q}{4C}=\frac{1}{4} \frac{Q}{C}=\frac{V}{4}$

which means

$V'=\frac{12.0 V}{4}=3.0 V$

3 0

1 year ago

Suppose a rectangular piece of aluminum has a length D, and its square cross section has the dimensions W XW, where D (W x W) to

Ludmilka [50]

Answer:

R₂ / R₁ = D / L

Explanation:

The resistance of a metal is

R = ρ L / A

Where ρ is the resistivity of aluminum, L is the length of the resistance and A its cross section

We apply this formal to both configurations

Small face measurements (W W)

The length is

L = W

Area

A = W W = W²

R₁ = ρ W / W² = ρ / W

Large face measurements (D L)

Length L = D= 2W

Area A = W L

R₂ = ρ D / WL = ρ 2W / W L = 2 ρ/L

The relationship is

R₂ / R₁ = 2W²/L

6 0

2 years ago

A 25.0-meter length of platinum wire with a cross-sectional area of 3.50 × 10^−6 meter^2 has a resistance

Nookie1986 [14]

R= (rou * L) / area
where R is the wire resistance
rou: resistivity of the wire material
L : wire length
A : cross section area of wire
by sub.
0.757= (rou*25)/ 3.5*10^-6
25*rou = 2.6495*10^-6
rou= 1.0598*10^-7 ohm.m

4 0

1 year ago